Optics & Modern Physics

Solutions of

Optics & Modern Physics

Lesson 26th to 30th

By DC Pandey

26. Reflection of Light Introductory Exercise 26.1 1 where c is the speed of light m 0 e0 1 in vacuum hence unit of is m/s. m 0 e0

w = 1.5 ´ 1011 rad/s

1. Since c =

2. Hence -7

11

By = 2 ´ 10 T sin [500x + 1.5 ´ 10 t ] Comparing this equation with the standard wave eqution By = B0 sin [ kx + w t ] 2p k = 500 m -1 Þ k = l 2p m Þ l= k 2p p m= metre Þ l= 500 250

Þ Þ

2 pn = 1.5 ´ 1011 1.5 n= ´ 1011 Hz 2p

Speed of the wave v =

w 1.5 ´ 1011 = k 500

= 3 ´ 108 m/s Let E0 be the amplitude of electric field. Then E0 = cB0 = 3 ´ 108 ´ 2 ´ 10-7 = 60 V/m Since wave is propagating along x-axis and B along y-axis, hence E must be along z-axis Þ E = 60 V/m sin [500x + 1.5 ´ 1011 t ]

Introductory Exercise 26.2 1. Total deviation produced

From figure q = 90° - i

1

Þ i i 90° 90°

d = 360° - 2 [ i + 90 - i] = 180°

2

180

°–2 i

N1

90°

q q

2. v0 = 2 m/s for plane mirror vi = 2 m/s. 180°–2q

N2

d = 180° - 2 i + 180° - 2q d = 360° - 2( i + q)

Hence rays 1 and 2 are parallel (antiparallel).

Velocity of approach = v0 + vi = 4 m/s. 3. In figure, AB is mirror, G is ground, CD is pole and M is the man. The minimum height to see the image of top of pole is = EN

2

8m

L' q

L

tan f =

A q

N

K

N1

M

C

f

2m

f f

B C' 6 m

f = 45° So, NK = 4 ´ tan 45° = 4 m Hence in minimum height = 6 m + 4 m = 10 m In

DAC¢ C

pole =4m E

2m

D

tan q = 2m

C

= EK + KN = 6 + KN Now in DNKB, NK = tan f Þ NK = KB tan f KB

4 =2 2

In DL¢ LA we get, LL¢ = tan q LA LL¢ Þ =2 4 LL¢ = 8 m

Þ

= 4 tan f

BC¢ 2 = =1 CC¢ 2

Maximum height = CA + LL¢ = 8 + 8 = 16 m

In DBC¢ C we get,

Introductory Exercise 26.3 1. Here f = - 10 cm (concave mirror) (a) u = - 25 cm Using mirror formula, 1 1 1 + = v u f Þ Þ Þ

1 1 1 1 1 = - =+ v f u 10 25 1 -5 + 2 = v 50 50 v== - 167 . cm 3

Hence image is real, inverted and less height of the object. (b) Since u = - 10 cm, Hence object is situated on focus of the image formed at ¥. (c) u - 5, f = - 10 1 1 1 1 1 = - =+ v f u 10 5

Þ

1 -1 + 2 = v 10

Þ

v = 10 cm

Hence, image is virtual, erect and two time of the object. 1 2. Here u = - 3 m, f = - m, 2 we have, 1 1 1 (a) = v f u Þ

1 1 = -2 + v 3

Þ

v = - 0.6 m

As ball moves towards focus the image moves towards -¥ and image is real as the distance decreases by focal length image become virtual which moves from + ¥ to zero. (b) The image of the ball coincide with ball, when u = - R = - 1 m

3 Using h = ut + Þ

t=

1 2 gt 2

4. Since the incident rays and reflected rays are parallel to each other therefore mirror is plane mirror.

2´ 2 2h = g 9.8

5. Let us solve the first case :

= 0.639 s Similarly again images match at t = 0.78 s. 3. Since image is magnified, hence the mirror is concave. -v -v Here, m= Þ =5 u u v = -5u

Þ

q M q

B

q A

a a

a A'

40 cm

…(ii)

From Eqs.(i) and (ii), we get - (5 + x) = - 5x 4x = 5

Þ

x = 1.25

Hence mirror is placed at 1.25 m on right side of the object by mirror formula 1 1 1 + = , v u f we have 1 1 1 =f 6.25 1.25 Þ

f =

- 6.25 , 6

Hence R = 2 f Þ R = -

C

20 cm

Since image is formed at a distance 5 m from mirror

Þ

q F

…(i)

Let distance between mirror and object is x.

v = - (5 + x)

2q

20 cm

By applying the geometry we can prove that, 40 PA¢ = v = cm 3 Further, in triangles ABP and PA¢ B¢ we have, AB A¢ B¢ = 40 ( 40/ 3) \

Thus mirror is concave mirror of radius of curvature 2.08 m.

AB 2 = cm 3 3

Similary, we can solve other parts also. 6. Simply apply : 1 1 1 = = v u f I -v for lateral magnification. If = o u magnitfication is positive, image will be virtual. If magnification is negative, image will be real. and m =

6.25 = - 2.08 m 3

A¢ B¢ =

4

AIEEE Corner ¢

Subjective Questions (Level 1) Hence the images distance are 2 nb, where n = 1, 2, K . Ans.

1. Here v = 39.2 cm, hence v = - 39.2 cm and magnification m = 1 hi = ho = 4.85

Þ

Hence image is formed at 39.2 cm behind the mirror and height of image is = 4.85 cm.

5. Suppose mirror is rotated at angle q about its axis perpendicular to both the incident ray and normal as shown in figure y

2. From figure, angle of incident = 15° N

0°

i

Mirror

i

°

15

N

Incident ray

90° 1 50

Reflacted ray

R

I

x

q (a)

Horizontal 15°

y

15° IV

Let reflected ray makes an angle q with the horizontal, then

I

i–q q i–q

q + 15° + 15° = 90° Þ q = 60° A

3.

R' i–2q

C

x

q (b) O'

1o cm

30 cm 50 cm

In figure (b) I remain unchanged N and R shift to N¢ and R¢.

50 cm B

O'''

1o cm

O''

30 cm

O'''

70 cm

From figure (a) angle of rotation = i,

D

40 cm

From figure (b) it is i - 2q

Since mirror are parallel to each other ¥ image are formed the distance of five closet to object are 20 cm, 60 cm, 80 cm, 100 cm and 140 cm.

Thus, reflected ray has been rotated by angle 2q. 6. I is incident ray Ði = 30° = Ðr

4. The distance of the object from images are 2b, 4 b, 6b..... etc.

30°

C

30°

A

1.6m

A

2b

B

20 cm R

O'''

b

O''

b

O'

b

O'''

I A'

4b

x

P

4b

1

From D PA¢ A, we get B

D

B'

5 x = tan 30° Þ x = 20 tan 30° 20 160 cm AB No. of reflection = = x 20 cm ´ tan 30° = 8 3 » 14 Hence the reflected ray reach other end after 14 reflections. 7. The deviation produced by mirror M1 is = 180° - 2 a M1 Z' I1 a 180°–2a R2 90°–a a R1 f f 90°–f 180°–2q q A

C

and the deviation produced by mirror M2 is = 180 - 2 Hence total deviation = 180 - 2a + 180 - 2f = 360 - 2 ( a + f)

Using mirror formula, Þ

1 1 1 = v f u

Þ

- 165 + 11 1 1 1 =+ = v 11 16.5 16.5 ´ 11

Þ

v=-

Hence size of image is hi = 2 ´ h0 = 2 ´ 6 = 12 mm. R 9. Here u = - 12 cm, f = + = + 10 cm 2 Using mirror formula 1 1 1 + = v u f we get 1 1 1 1 1 = - = + v f u 10 12

90 - a + q + 90 - f = 180

Object height h0 = 6 mm u = - 16.5 cm (a) The ray diagram is shown in figure B

Þ

The image is formed on right side of the 60 vertex at a distance cm. the image is 11 virtual and erect the absolute magnification v is given by |m|=½ ½ ½ u½ Þ

A' A

f

B' u = 16.5 cm

6+5 60 60 cm = 5.46 cm v= 11 =

a + f=q

Hence deviation produces = 180 - 2q. R 22 8. Here f = - = = - 11 cm 2 2

16.5 ´ 11 = - 33 cm 5.5

Hence the image is formed at 33 cm from the pole (vertex) of mirror on the object side the image is real, inverted and magnified. The absolute magnification v 33 |m|=½ ½ = =2 ½ u½ 16.5

In D ABC we get,

Þ

1 1 1 + = v u f

Q

½ ½ 5 60 ½= |m|=½ ½ 11 ´ ( - 12) ½ 11 m 1. Therefore mirror is concave and Q m is -ve. Hence image is real [for concave mirror m is = - ve]

(b) ® r (c) ® p 3. (a) Since object and its image are on opposite side of principle axis. O

Therefore, A

(a) ® q, r 1 (b) Since m = - , Q m is - ve 2 Hence mirror is concave and image is real. (b) ® q, r (c) m = + 2 , Q m > 1 Hence mirror is concave and Q m is + ve

B I

Hence mirror is concave Þ (a) ® r. (b) Similarly as for option (a). (b) ® r

Hence image is virtual.

(c) Since image and object are of same height from AB.

(c) ® q, s

Hence mirror is plane mirror.

1 (d) Q1 m = + < 1 and + ve 2

(c) ® p

Hence the mirror is convex and image is virtual. (d) ® p, s 2. Plane mirror (for virtual object) ® only real image Þ (a) ® p

(d) Since image is magnified. I

O x A

x B

Hence mirror is concave [D is. distance between O and mirror is less than the focal length].

18 Hence

m =2 = -

(d) ® r. 4. (a) For concave mirror M1 focal length = - 20 cm

Using

1 1 1 - = u u f

we get,

-

Þ

3 1 = 2 u 20

Þ

u = 30 cm

When x = 20 cm, Mirror is M1 v = ¥ and magnified (a) ® p, s (b) For convex mirror M2 of focal length + 20 cm if X (distance of object from pole) = 20 1 1 1 Using mirror formula + = v v f

1 1 1 - =2u u 20

If image is virtual v = 2 v 1 1 1 - =2u u 20 u = 10 cm

Þ

Hence correct option are as

we get 1 1 1 1 = + = v 20 20 10 v = 10 cm

Þ

v Þ v = - 2u u

Hence image is virtual. (b) ® r (c) u = - 30 cm, f = - 20 cm 1 1 1 2 -3 1 = = =v 30 20 60 60

(a) ® p, q (b) Here m =

Hence image is real. 1 v u Þ = - ®v = 2 u 2 1 1 1 Using + = , we get v u f -

v = - 60 cm Hence image is real. 60 m== -2 30 Hence image is magnified (2 times). Þ (c) ® q, s (d) for mirror M2 (convex) at X = + 30 cm image again virtual. (d) ® r 5. (a) For concave mirror f = - 20 cm Case I. Image is real.

1 53 ° Þ

Þ

x = 20 cm

Hence answer is (c).

é 2 ù Dt = t ê1 ú êë wm g ú û é 3/2 9 1 ù = 36 ê1 = úQ wm g = 4/3 8 9/8 û ë 36 ´ 1 = = 4 cm 9 Hence correct option is (b). Real depth 22. m = App. depth Þ

Þ

1 1 1 + = OB OA f

Þ

OB. OA Q OB + OA = AB f = OA + OB

4 real depth = 3 10.5 cm

Þ Real depth =

4 ´ 10.5 cm = 14 cm 3

Hence correct option is (d). 23. Q y0 = + 1 cm and yi = - 2 cm

19. The ratio of focal length in the situation II and III is 1 : 1. Hence correct option is (c). 1 1 1 20. We have = OB ( - OA) f

OC2 AB

21. The shift produce

Hence correct option is (a).

Þ

…(ii)

Hence correct option is (c).

q < 37 °

This image act as virtual object for plane-glass-water surface m g m w ( x - 20) m g - m w Þ = ¥ 20x ¥

f =

Þ

90 + q + i = 180°

18. For reflection at curved surface 1 1 3 1 = æç - 1 ö÷ ´ v ( -x) è 2 ø 10 20x Þ v= x - 20

OC2 = OA OB

Putting this value in Eq. (i), we get

q = 90 - iQ i > 53 ° Þ

…(i)

Þ OA2 + OB2 + 2 OAOB = 20 C2 + OA2 + OB2 C

sin i =

OB. OA AB

(OA + OB)2 = OC2 + OA2 + OB2 + OC2

sin i 6/5 = sin 90 3/2 Þ

f =

Now AB2 = AC2 + BC2

6/5

q A

Þ

m=-

v v Þ2= u u

Now let x be the position of lens then v = 50 - x and v = ( 40 + x). 50 - x Þ 2= 40 + x Þ

80 + 2x = 50 - x

Þ

- 3 x = 30 Þ x = - 10 cm

Hence correct option is (c).

41 24. If the plane surface of plano-convex lens is silvered it behave the concave mirror of focal length fm /2 Q

fm = 10 cm

Þ

fe = 5 cm hence R = 10 cm

Correct option is (c).

and

öæ 1 1 æç m w 1 = - 1 ÷ çç + ç ÷ f èmg ø è R1 R2

ö ÷÷ ø

…(ii)

Dividing (i) and (ii) m w lg 4 4 = Þ = m g lw 3mg 5 Þ

25. Since lens made real and magnified image, hence it is a convex lens when lens dipped in water its focal length.

c = n lw mw

mg =

5 3

Hence correct option is (a). f f + d ( f1 - d) ( f - d) D 29. x = 1 2 and y = 1 f1 + f2 - d f1 + f2 - d f = 20 cm

f = – 20

æ 4/3 öæ 1 1 ö -1 æ 1 1 ö ÷÷ = çç ÷÷ = çç - 1 ÷÷ çç + + 3 / 2 R R 9 R R 2 ø 2 ø è 1 è øè 1 Q f is - ve lens behave as concave, hence the image is virtual and magnified. Correct option is (c).

5 mm

30 cm

Here f1 = f2 = 20 cm , d = 30 cm and D = 55m = 0.5 cm

26. The prism transmit the light for which angle of incidence(c), 2c £ 90° Þ c £ 45°

Putting these values we get x = 25 cm and y = 0.25 cm Correct option is (b).

90°

Hence m =

1 1 = = 2 = 1.414 sin C sin 45°

Correct option is (b). 1 1 1 v 27. = Q|m|= =3 v ( -16) f u For convex lens v = 3u 1 1 1 + = Þ f = 12 cm (for real image) 48 16 f Similarly when distance is 6 cm, 3 times virtual image is formed hence mirror is convex with focal length 12 cm. Correct option is (c). c 28. v g = nl g Þ = nl g …(i) mg

30. Since for each q angle of incidence at glass-air boundry remains 0° hence there will never be total internal reflection. Correct option is (d). 31. Diameter = m ´ Original diameter =

4 4 ´ 1 cm = - cm 3 3

Hence correct option is (a). 1 1 1ù 32. = (1.5 - 1) é + êë20 20 úû f 1

Þ

f1 = 20 cm

Here u1 = - 30 cm 1 1 1 1 1 1 Þ v1 = 60 cm = Þ = v u1 f1 v1 20 30 Magnification|m1|= (Inverted image)

v 60 = =+2 u 30

42 For second lens.

1 1 1 Þ v2 = + 30 cm + = v2 60 20

Magnification m2 =

30 1 = 60 2

Total magnification = m1 ´ m2 = 1 Hence object size remains 3mm and it is formed at (120 + 30) cm = 50 cm from first lens.

Þ

Hence correct option is (d). 1 ö 36. Shift in mirror = 6 æç 1 ÷ = 2 cm 1.5 ø è 6 cm

Hence correct option is (b).

Man

33. From left hand side refraction occur from n2 = 2 to n1 = 1. n 1 n = 1 = = 0.5 n2 2 n 1 n -1 - = v u R 0.5 1 -0.5 1 0.5 = + = v 10 10 10 10 Þ

v = 10 cm

1 1 1 =v 20 30 1 1 1 Þ v = + 60 cm = v 20 30

50 cm

Hence man lie at 48 cm from mirror. The distance of image from observer = 2 ´ 48 = 96 cm Hence correct option is (b). 1 1 1 37. Using lens formula - = v u f

Hence correct option is (a).

1 1 1 + = ( f + 40) ( f + 10) f

Þ

34. For lens v = - 20 cm, f = + 10 cm Using

On solving we get f = 20 cm

1 1 1 - = we get v = 20 cm v u f

This image acts virtual object for convex mirror. For mirror v = - ( x - 20)

1.5

Hence correct option is (c). 38.

q

and v = - (20 + x) and f = + 60 cm 1 1 1 Using + = v u f

q

sin q = n1 sin (90 - q c )

1 1 1 = - (20 + x) ( x - 20) 60

sin q = n1 cos q c

After solving we get x = 20 cm. Hence correct option is (c). 35. In this case system behave as concave mirror or focal length = m ´ fe

Now

Þ sin q <

= 1.5 ´ 20 = 30 cm Now using mirror formula 1 1 1 + = we get v u f

Þ

cos q c <

n12 - n22 n1

n12 - n22 ´ n1 n1 sin q < n12 - n22

Hence correct option is (a).

43 39. For

mirror u = - 1 cm

(taking

1 1 1 0.6 0.3 = + = + F f1 f2 10 20 ´ 1.6

upward

direction + ve) f = - 2 cm 1 1 1 =- v f u Þ

1 1 1 =- +1= v 2 2

Þ

v = 2 cm

Hence mirror form virtual image behind mirror at 2 cm from pole. This image acts as virtual object for slab on see below the slab the shift is æ 1 ö ÷ ´ 9 = 3 cm Dt = çç 1 3/2 ÷ø è Hence virtual image form on object thus correct option is (a). m - m1 m m 40. We have 2 - 1 = 2 v u R For real object u is + ve m 2 (m 2 - m 1) m 1 Þ = + v R u Q Þ Þ

R is + ve and m 1 > m 2 m2 m = - ve + 1 v u v = - ve

Hence, if m 1 > m 2 then these cannot be real image of real object. Hence, correct option is (a). 41. At oil-concave surface 1 1 1 ù 1 0.6 Þ = = (1.6 - 1) é + êë ¥ 10 úû f f1 10 1 At other surface light goes from oil to glass 1 æ 1.5 1 1 =ç - 1 ö÷ é- ù ê f2 è 1.6 ø ë 10 20 úû 1 0.3 = f2 20 ´ 1.6 Let focal length of combination is F

Þ

F = 28.57 cm

Hence correct option is (d). 42. Using formula m2 m1 m2 - m1 = v u R For real image m is -ve m2 m1 m2 - m1 Þ + = v u R 1.5 0.5 1 Þ = v R u 1.5 1 1 Þ = - v is + ve if u > 2R v 2R u Hence correct option is (b). fu 43. v = when u < f f -u u lend to lens Þ if

u ® 0, v ® 0

Hence correct option is (d). 44. Since image formed by diverging lens is always virtual. Hence correct option is (a) 45. If the object place at first focus the image forms at ¥ ie, rays incident the plane surface normally and retrace its path. - 60 -m 1R Now x = f1 = = - 120 cm m 2 - m 1 0.5 Hence correct option is (a). m - m1 m m 46. 2 - 1 = 2 for m 2 = 1.6 v1 u1 R 1.6 - 1 6 1.6 1 = = v1 ( -2) 1 10 Þ

1.6 6 1 1 = - = v1 10 2 10

Þ

v1 = 16 m

For

m2 = 2

44 2 -1 2 1 = =1 v2 ( -2) 1

Þ

Hence correct option is (a).

2 1 1 = 1 - = Þ v2 = 4 m v2 2 2

48. The focal length of lens combination is 2f

Hence separation between images = v1 - v2

Hence

= (16 - 4) cm

1 (1.8 - 1.2) é 1 1 ù = + êë ¥ R úû 2f (1.2 - 1) 1 0.6 = 16 0.2R

= 12 cm Hence correct option is (a).

Þ

47. System behave a a concave mirror

R = 48 cm

Hence correct option is (a).

Here u = - 10, v = - 40 cm

49. If plane surface is silvered the system acts a concave mirror having focal length R = = 24 cm 2

Using mirror formula 1 1 1 + = we get v u f -1 1 1 = 40 10 f

¢

f = - 8 cm

Hence correct option is (c).

More than one options correct

1. Since prism are identical hence if a right prism produce deviation d inverted prism produce deviation -d. if n = 2m deviation becomes zero.

The correct options are (b) and (d). 3. The correct options are (b) and (c). 4. The lens form real image if D ³ 4 F (displacement method)

if n = 2m + 1 deviation produce is d Hence correct options are (a) and (b). 2. sin i

and

f =

D2 - x2 4D

and the magnification m1m2 = 1 Hence correct options are (b), (c) and (d). 5. Deviation produced by prism 30°

d = (m - 1) A = (1.5 - 1) 4 ° = 2° sin r

if the mirror is rotated q = 2° ray become horizontal after reflection from mirror.

Q sin i = m sin r Þ

m = tan 30° = 3

if speed of light in medium x is v then speed of light; in medium y =

v v = m 3

Since y is denser w.r.t. x hence total internal reflection take place when incidence in y.

Again if mirror is rotated by 1° reflection ray deviated by 2° from horizontal and after passing through prism again ray become horizontal. Hence correct option are (a) and (b).

45 ¢

Match the Columns

1. Correct match is (a) ® q, r (c) ® p, r

(b) ® p, s (d) ® p, r

4. (a) ® q,

2. (a) ® p, r

(b) ® q, s

(c) ® q, r

(d) ® q, r

(b) ® r (c) ® r (d) ® p 5. Since

O

3.

m1

here n = 1.5 for (a) n = 1.4

x 1

I m2

2

é1 1 1 ù = ( n - 1) ê ú f ë R1 R2 û

m2 > m

Þ

æ 1 1 æ 1.5 1 ö ÷ =ç - 1 ö÷ çç f è 1.4 ø è R1 R2 ÷ø

Þ

f is + ve

Hence image distance is less than x and virtual.

hence

Þ (a) ® q, s Similarly as in (a) the correct match for

(b) ® q, s Q f is - ve and increase in magnification

(b) ® q, r

Similarly

(c) ® 1

(c) ® q, s and

(a) > p, s Q f increases, power decreases

(d) ® p, s 2

O

(c) ® p, s (d) p, r

6. For real object at 2c convex lens form image at 2c similarly for virtual object concave lens does. (a) ® q, s (b) ® q, r (c) ® q, r (d) ® q, r

28 Interference and Diffraction of Light Introductory Exercise 28.1 1. Because they are incoherent ie, Df does not remain constant. 2. Since laser is highly coherent monochromatic source of light

and

3. I = I 0 cos2 q / 2 Þ

3I0 = I 0 cos2 q / 2 4

Þ

cos q / 2 = q p = 2 6 p q= 3

Þ Þ Þ

l l p l ´ f= ´ = 2l 2p 3 6 yd Dx = D D Dx D l y= = ´ d d 6 Dx =

But Þ

3 2

1.2 ´ 600 ´ 10-9 Þy= = 48 mm 0 . 25 ´ 10-2 ´ 6 1 4. 2 mt = æç n - ö÷ l for minimum thickness n = 1 2ø è l 3 Þ t= = = 0.5 cm 4m 4 ´ 1.5 5. Here a1 = 3 a and a2 = a R2 = (3 a)2 + ( a)2 + 2 ´ 3 a ´ a ´ cos q Þ I = 9 I 0 + I 0 + 6I 0 cos q Þ I = [10 + 6 cos q ] I 0

é2 cos2 q - 1ù ù I êë úû úû 0 2 q = é10 + 12 cos2 - 6ù I 0 êë úû 2 q = é4 + 12 cos2 ù I 0 êë 2 úû q = 4 I 0 é1 + 3 cos2 ù êë 2 úû I Now, I max = 0 9 4 q Þ I = I max é1 + 3 cos2 ù ê 9 2 úû ë yd 6. Dx = - (m - 1) t D l if t = 2 (m - 1) é I = ê10 + 6 ´ ë

Þ Dx =

yd l D 2

For maxima Dx = nl yd æ 1 Þ = ç 2n + ö÷ l D è 2ø This become minima. 1 For minima Dx = æç n - ö÷ l 2ø è yd l l - = nl D 2 2 yd Þ - nl this become maxima. D Hence maxima interchanged.

and

minima

are

47 7. For two slit experiment

Þ But

8. Since amplitude of each wave is equal. The amplitude of resultant wave is zero if waves are equally displaced in phase 360° ie, q= = 45° 8

d sin q = nl nl sin q = d nl sin q £ 1 Þ £1 d n£

Þ

n =6

Þ

4 ´ 10-6 d Þ n£ = 6.67 l 6 ´ 10-7

Hence phase difference must be 45°

AIEEE Corner ¢

Subjective Question (Level-1)

1. R2 = a12 + a22 + 2a1a2 cos f

Þ

(i) R = 2a, a1 = a2 = a 4 a2 = a2 + a2 + 2a2 cos f

Now,

Þ cos f = 1 Þ f = 0° (ii) 2a2 = 2a2 + 2a2 cos f Þ

Þ f = 90° (iii) a2 = 2a2 + 2a2 cos f -1 Þ cos f = Þ f = 120° 2

But

(iv) 0 = 2a2 + 2a2 cos f Þ f = 180° 2

2.

I max I min

æ a1 ö çç + 1 ÷÷ 2 a ( a + a2 ) ø = 1 =è 2 2 ( a1 - a2 )2 æ a ö çç 1 - 1 ÷÷ è a2 ø 2

Þ

I max I min

æ 5 ö ç ÷ 2 ç3 + 1÷ è ø = 8 = 16 = 2 22 æ 5 ö ç ÷ ç3 -1÷ è ø

I max : I min = 16 : 1 q 3. I = I max cos2 2 I max q Þ = I max cos2 2 2 q p Þ cos = cos 2 4

Þ Þ

p 2 2p f= ´ Dx l p 2p = ´ Dx 2 l l Dx = 4 yd Dx = D D Dx y= d q=

y=

1m ´ l 1 ´ 500 ´ 10-9 = 4 ´ 1 mm 4 ´ 10-3

y = 1.25 ´ 10-4 m q 4. I = I max cos2 2 I q p (a) max = I max cos2 Þ q = 2 2 2 2p Now q = ´ Dx l l yd Dl Þ Dx = and Dx = Þ = 4 D 4d 1 2 q (b) I max = I max cos 4 2 q p 2p Þ cos = cos Þ q= 2 3 3 2p f= ´ Dx l Þ

48 2p 2 p l = ´ Dx Þ Dx = 3 l 3 yd Dl Now Dx = Þy= D 3d q 5. (a) I = I max cos2 2

8. In Young’s double slit experiment Dl w= d 2.2 ´ l Þ 2.82 ´ 10-3m = .460 ´ 10-3

Þ

I max = I 0 and q = 60° 3 Þ I = I 0 cos2 30° = I 0 = 0.75I 0 4 2p (b) q = ´ Dx l p 2p Þ = ´ Dx 3 l l 480 nm Þ Dx = = 6 6

9. xn for bright fringe is given by nDl d Dl 2Dl and x2 = x1 = d d Dl the angular separation Dx = d xn = Þ

sin ( Dq) =

Dx = 80 nm

Þ

l = 590 nm

Þ

6. l A = lB = 6m

( Dq) = sin -1 [0.00025]

A

B

= 0.014 °

5m

For constructive interference difference wavelength = 0, l, 2l Q

Dx l 5 ´ 10-7 = = = 2.5 ´ 10-4 D d 2 ´ 10-3

l = 6 > dAB = 5

10. When whole appratus is immersed in water 4 l = nw l = ´ 700 ´ 10-9 m 3 w=

Hence only constructive interference occur at Dl = 0 5 Þ x = m = 2.5 m 2 3l For destructive interference Dl = l, 2

Dl 48 ´ 10-2 ´ 4 ´ 7 ´ 10-7 = = 0.90 mm d 3 ´ 25 ´ 10-5

D1l Dl and w2 = 2 d d ( D1 - D2 ) l Þ w1 - w2 = d

11. w1 =

Þ 3 ´ 10-5 =

Only possibility at Dl = 6 Which occur at x = 1 m and x = 4 m from A. 7. The wavelength l = A

3 ´ 108 c = = 2.5 m n 120 ´ 106 (9 – x)

B

x

For constructive interference x - (9 - x) = nl , where n = 0, 1, 2 K x = 4.5, 5.75, 7, 8.25 The other points are 3.25, 2, .75

1.5 ´ 10-2 ´l 10-3

3 10-8 ´ -2 = 2 ´ 10-6 m = 2 mm 1.5 10 D nl 12. For bright fringe xn = d Þ l=

For first light ( l = 480 nm ) the third order 1 ´ 3 ´ 480 ´ 10-9 Bright fringe is x3 = 5 ´ 10-3 For second light ( l = 600 nm ) x¢3 =

1 ´ 3 ´ 600 ´ 10-9 5 ´ 10-3

49 Dx = x¢3 - x3 =

3 ´ 10-9 ´ (600 - 480) 5 ´ 10-3

=

3 ´ 120 ´ 10-6 5 -6

Dx = 72 ´ 10 , Dx = 72 mm 13. Fringe width : w= =

lD d

(500 ´ 10-9 ) (75 ´ 10-2 ) m (0.45 ´ 10-3)

= 0.83 ´ 10-3 m = 0.83 mm Distance between second and third dark line = one fringe width = 0.83 mm. 14. For first order bright fringe Dl x= d Þ

4.94 ´ 10-3 =

3 ´ 600 ´ 10 d

= 3l I

15.

y2 S1

2

D +

y12

=

3l 2d

y1

Þ

(35 ´ 10-2 )2 + y12 y2 (35 ´ 10-2 )2 + y22

=

550 ´ 10-9 2 ´ 1.8 ´ 10-6

=

…(i)

3 ´ 550 ´ 10-9 2 ´ 1.8 ´ 10-6

l

16. Wavelength of source l =

2 meV -34

Þ l=

6.62 ´ 10

2 ´ 9.1 ´ 10-31 ´ 100 ´ 1.6 ´ 10-19

Þ l = 1.24 ´ 10-10 m w=

Dl 3 ´ 1.24 ´ 10-10 = m = 36.6 cm d 10 ´ 10-10

17. Given l = 546 nm = 5.46 ´ 10-7 m, D = 1m

(a) At distance y = 10 mm = 10 ´ 10-3m from the central fringe path difference will be Dy =

screen

y. d 10 ´ 10-3 ´ 0.3 ´ 10-3 = D 1

= 3 ´ 10-6 m

y1 q q¢

S2

y2

and

d = 0.3 mm = 0.3 ´ 10-3m

l = 12 ´ 10-7 m l = 1200 nm

Þ

For first dark fringe n = 1 and for 2nd, n = 2 l dsin q = 2 3l l Þ sin q = dsin q¢ = 2 2d y1 l Þ = D2 + y12 2d

On solving y2 - y1 = 12.6 cm

obtained at this point for first dark fringe Dl x= 2d 3 ´ l ´ 4.94 Þ 4.94 ´ 10-3 = 2 ´ 18 ´ 10-4 Þ 2 ´ 18 ´ 10

l 2

-9

Let for wavelength l first dark fringe is

-7

d sin q = (2n - 1)

and

18 ´ 10-7 18 d= = ´ 10-4 m 4.94 ´ 10-3 5.94

Þ

For dark fringe

Corresponding phase difference 2l 2p f= ´ Dx = ´ 3 ´ 10-6 rad l 5.46 ´ 10-7 = 1978 °

50 I = I 0 cos2

q 2

= I 0 cos2 æç è

Þ 1978 ° ö 2 ÷ = I 0 cos 989 ° 2 ø

I = 3 ´ 10-4 I 0 (b) Fringe width w = w=

Number of fringes =

w=

10 mm = 5.49 1.82 mm

Hence the number of fringe is five. 18. Shift due to sheet of thickness 10m and

1.1784 D ´ 10-6 m d

Now when t is removed and D is doubled the distance between successive maximum (or minima) i. e., fringe width

lD d

5.46 ´ 10-7 ´ 1 = 1.82 mm 0.3 ´ 10-3

2Dl d

but according to question Dx = w Þ Þ

1.178 ´ 10-6 D 2 Dl = d d l = 0.589 ´ 10-6 m = 589 nm

20. Let n bright fringe (l = 5500 Å) concide with 10th bright fringe of 6000 Å

refractive index 1.6 is (m - 1) tD Dx1 = d

n ´ 5500 Å = 6000 Å ´ 10

Þ

10 ´ 10-6 ´ 1.5 Þ Dx1 = (1.6 - 1) ´ 1.5 ´ 10-3 Þ Dx1 = 0.6 ´ 10 ´ 10-3 = 6 ´ 10-3m Shift due to sheet of thickness 15m and refractive index 1.2 is -6

Dx2 =

Dx =

(1.2 - 1) ´ 15 ´ 10 1.5 ´ 10-3

n~ - 11

Þ

Similarly first bright fringe concide with 1st fringe. Now fringe width 14.74 - 12.5 w= = 0.224 mm 10 Hence position of 10th bright fringe

´ 1.5

= 14.74 - 0 . 224 ~ - 14.55 mm Position of zero order bright fringe

Dx2 = 3 ´ 10-3m

= 12.75 - 0.224 ~ - 12.25 mm

Since these shifts are in opposite direction of central maxima hence net shift

21. Here d ~ - 1 cm

Dx = Dx1 - Dx2 = 6 ´ 10-3m - 3 ´ 10-3m P

= 3 ´ 10-3m = 3 mm 19. Let l is the wavelength of light D is screen distance from source and d is the separation between slits (all are in metres) Shift = Dx = Þ

Dx =

S

y

1 cm q

(m - 1) tD d

(1.6 - 1) ´ 1.964 ´ 10-6 ´ D d

D = 100 m l = 500 nm

51 For first dark fringe -9

y=

Dl 100 ´ 500 ´ 10 = 2d 2 ´ 10-2

y = 2.5 mm

Þ

22. The destructive interference will be 2mt = nl for thinnest n = 1

25. For destructive interference 2mt = nl for t to be minimum

Þ

2mt = l

Þ

t=

Þ

t = 114 nm

l 650 ´ 10-9 = 2m 2 ´ 1.42

23. Here t = 0.485 mm , t = 485 nm, n = 1.53 (a) Condition for constructive interference in refractive system 1 2mt = æç n - ö÷ l where, n = 1, 2 2ø è For

24. For constructive interference 1 2mt = æç n - ö÷ l for t to be minimum n = 1 2ø è l 6000 Þ t= = = 1154 Å 4m 4 ´ 1.3

n = 1, l1 = 4mt = 4 ´ 1.53 ´ 485 = 2968.2 mm

Þ

n =1 l 800 t= = 2m 2m

For constructive interference l 2mt = (2n - 1) 2 800 l 2m ´ = (2n - 1) 2m 2 1600 = (2n - 1) l For n = 1, l = 600 nm which does not lie in visible region

which does not lie in visible region put

For

n = 2, 3, ……

2mt = nl, putting n = 1, 2,

1600 = 3 l 1600 Þ = l Þ l = 533 nm 3 1600 for n = 3, l = = 320 nm which does lie 5 in visible

only l = 495 nm is lie in visible region.

Hence l = 533 nm

we get l = 424 nm, 594 nm, … (b) For structive interference in transmitted system

n =2

Objective Questions (Level-1) ¢

…(i)

(Single option correct) R = 82 + 62 = 10 m

1. Using phasor method 6m

Hence correct option is (b). 2. Here

8m

12 m

I1 = B2 I2 I max = [ I1 + I1 ]2

6m

I min = [ I1 - I2 ]2 8m

52

I min I max

Þ

é ê =ê ê ê ë

2

( I max - I min ) éb + 1 - b + 1 ù = ú ( I max + I min ) êë 2b û

Þ

(m - 1) tD d (1.5 - 1) ´ 10 ´ 10-6 ´ 100 ´ 10-2 = 2.5 ´ 10-3

ù I1 - 1ú 2 I2 ú = éb - 1 ù ê ú ú I1 ëb + 1 û + 1ú I2 û

5. Shift =

=

2

Hence correct option is (a). Dnl 6. xn = for nth bright fringe d Dl xn ¢ = (2n - 1) for nth dark fringe 2d Dl Dx = xn - xn ¢ = [2n - 2n + 1] 2d Dl Dx = 2d

I max - I min 1 = I max - I min b2

Hence, correct option is (d). 3. For nth dark fringe xn = (2n - 1)

Dl 2d

For 1st dark fringe n = 1

Hence correct option is (c).

Dl x1 = 2d

Þ

5 ´ 10-6 = 2 ´ 10-3m = 2 mm 2.5 ´ 10-3

7. Since at centre path difference for all colour is always zero hence centre will be white.

x Angular position q = sin -1 æç 1 ö÷ è Dø l Þ q = sin -1 é ù êë2a úû Þ

é 5460 ´ 10-10 ù q = sin -1 ê -3 ú ë2 ´ 0.1 ´ 10 û

Þ

q = sin -1 [273 ´ 10-5 ] -1

Þ

q = sin

Þ

q = 0.16°

Hence correct option is (a).

[0.00273 ]

8. n1 ´ l1 = n2 ´ l2 60 ´ 4000 = n2 ´ 6000 Þ n2 = 40

Hence correct option is (b) 4. For 10th bright fringe x10

10Dl = d

For 6th dark fringe x6 = (2 ´ 6 - 1) But

Þ But

Dl¢ 2d

x6 = x10 10 Dl 11 Dl¢ = 2a d l¢ 20 = l 11 l¢ 20 m= = = 1.8 l 11

Hence correct option is (a).

Hence correct option is (a). Dl 9. Initial fringe width w1 = d Final fringe width w2 =

( D - 5 ´ 10-2 )l d

l ´ 5 ´ 10-2 d l = -3 ´ 5 ´ 10-2 10

|Dw|=|w2 - w1|= Þ

3 ´ 10-5

Þ

3 ´ 10-6 = l Þ 6000 Å = l 5

Hence correct option is (a).

53 10.

I max 49 = I min 9 é I + I2 ù I Now max = ê 1 ú I min êë I1 - I2 úû é I max ê =ê I min ê êë I 25 Þ max = I min 4

2

é ê =ê ê ê ë

I1 I2 I1 I2

ù + 1ú ú ú - 1ú û

2

2

2 ù 49 é7 + 1ù + 1ú ê ú 100 9 ú = ê3 ú = 16 7 49 ê - 1ú -1ú ë3 û úû 9

¢

l=

Hence correct option is (b). 12. Let nth fringe of 6500 Å concide with n¢th fringe of 5200 Å. Dn ´ 6500 Å Dn¢ = ´ 5200 Å d d n 5200 4 Þ = = n¢ 6500 5 Þ xn =

It means 4th fringe of 6500 Å coincide with 5th fringe of 5200 Å hence the distance x=

Hence correct option is (a) Dnl 11. xn = d 3 Dl Þ x3 = = 7.5 ´ 10-3m d Þ

l = 0.5 ´ 10-6 = 500 nm

Þ

7.5 ´ 10-3 ´ 0.2 ´ 10-3 3´1

4 ´ 120 ´ 10-2 ´ 6500 ´ 10-10 2 ´ 10-3

x = 0.156 cm Hence correct option is (a). 13. Since number of minima does not depends on orientation hence n1 = n2 Hence correct option is (a).

Assertion and Reason

1. We have I = 4 I 0 cos2

f 2p if f = 2 3

we get I = I 0 Hence assertion is true. l Now path difference = ´phase difference 2p l 2p l Path difference = ´ = 2p 3 3 Hence reason is true. But reason is not the explanation of assertion. Hence correct option is (b). 2. Here assertion and reason are both true but reason is not correct explanation of assertion. Correct option is (b).

3. Assertion is wrong since fringes are symmetrical ie, fringes obtained both above and below point O. Reason is true. Correct option is (a). 4. Here both assertion and reason are true and reason correctly explain assertion. Hence correct option is (a). 5. Both assertion and reason are true and reason correctly explain the assertion. Hence correct option is (a). 6. Assertion is true since locus of all fringes is circle. But reason is wrong since fringes may have any shape. Correct option is (c).

54 7.

9. Q d sin q = nl P

nl d but = 4 d l n Þ sin q = 4 Þ sin q =

d cos q q d S1 S2

Now if q = 30° n = 4 sin 30° = 2

Path difference at point P is Dx = d cos q

Þ

The path difference decreases as q increases. Q as q ® increases, cos q ® decreases d cos q Hence order of fringe n = decreases l as we go above P. Hence assertion is wrong (false).

Hence assertion is true.

For 11th order maxima path differenc is more hence reason is true but assertion is false correct option is (d). 8. Here assertion is true and reason is false and reason does not correctly explain assertion. Correct option is (c).

Also reason is true and does not correctly explain assertion correct option is (b). 10. Here assertion is false. Since (m - 1) tD is independent of l. = d

shift

Hence shift of red colour = shift of violet colour. and reason is true Q

mV >mR

Hence correct option is (d)

Objective Questions (Level 2) 1. I = 4 I 0 cos2

px a

k px = k cos2 æç ö÷ 4 è l ø px 1 Þ cos = l 2 px 1 = cos -1æç ö÷ l è2ø px p px 2p Þ = or = l 3 l 3 l 2l Þ x = or x = 3 3 Þ

lx1 a

Þ

I 0 = 4 I 0 cos2

Þ

x1 =

and

2I 0 = 4 I 0 cos2

Þ

x2 =

a 3

…(i) px2 a

a 4

Dx = x1 - x2 =

…(ii) a a a - = 3 4 12

Hence correct option is (c). px px 2. I = 4 I 0 cos2 æç ö÷ = k cos2 æç ö÷ è l ø è l ø p I = k cos2 æç ´ l ö÷ = k èl ø I p x = k cos2 æç ö÷ 4 è l ø

Hence correct option is (b). 3. Light of wavelength l is strongly reflected if 1 …(i) 2 ut = æç n + ö÷ l n = 0, 1, 2 2ø è …(i)

Þ 2 ut = 2 ´ 1.5 ´ 5 ´ 10-7 m = 1.5 ´ 10-6 m Putting l = 400 nm in eq. (1) and using eq. (ii)

55 1 1.5 ´ 10-6 m = æç n + ö÷ ´ 4 ´ 10-7 m 2ø è n = 3.25

Þ Putting -6

1.5 ´ 10

æ 3ö æ 3ö a ÷÷ Þ x = cos -1çç ÷÷ px = a cos -1çç 4 p è ø è 4ø x = 0.20 mm

Þ

l = 700 1 m = æç n + ö÷ ´ 7 ´ 10-7 2ø è

Hence correct option is (d). 6. Number of fringes shifted =

n = 1.66

4=

Hence n can take values 2 and 3. 4ut From (i) if n = 2 , l = = 600 nm 2´ 2 + 1 if

n = 3 l = 429 nm

only l = 600 is given in the options. Hence correct option is (b).

(m - 1) t l (1.5 - 1) ´ t

Þ

4 ´ 6 ´ 10-7 m = t 0.5

Þ

t = 4.8 mm

6000 Å

Hence correct option is (a). 7. For nth order minima

-2

4. xn =

Dln 100 ´ 10 = l´ n d 001 . ´ 10-3 nl xn = - 5 10

Þ

xn =

n ´ 4000 ´ 10-10 m = 4 ´ 10-5 ´ n 10-5

D

S2

(2n - 1) Dl for 3rd minima n = 3 2d 5Dl 5l y3 = = 2d 2q d Q q~ - tan q = D yn =

Þ xn = 4 n mm Similarly for l = 7000 Å xn = 7 n mm n = 5, 6 hence only x = 5 Passes through hole l = 5000 Å Hence correct option is (b). px Dl 5. I = 4 I 0 cos2 æç ö÷ where a = a d è ø -10

1 ´ 6000 ´ 10 1 ´ 10-3

3 px = cos 4 a

Hence correct option is (b). 3a ´ 5 8. AB = d = 3 a = 5 C

= (6 ´ 10-4 )

px 75% I 0 = I 0 cos2 æç ö÷ è a ø 3 px = cos2 16 a Þ

q q¢

For l = 4000 Å

= 0.04 ´ n

Þ a=

S1

(– a, 0) A

(2a, 0) D

AB = 15l Hence total maxima = 14 ´ 4 + 4 = 60° Hence correct option is (a).

B

56 px 9. I = 4 I 0 cos2 é ù êë a úû

= 12 ´ 104 ´ 10-9 m

p a Þ I = 4 I 0 cos2 é ´ ù êë a 4 úû p I = 4 I 0 cos2 4 4I0 = = 2I 0 2 I 1 Þ 0 = I 2 Hence correct option is (b). p 10. I = I 0 cos2 é (m - 1) t ù êë a úû at m = I Þ

I = I0

Hence correct variation is (c) px 11. I = I 0 cos2 æç ö÷ è l ø 3 px I 0 = I 0 cos2 æç ö÷ 4 è l ø px p Þ = l 6 l Þ x = but x = (m - 1) t 6 l Þ (1.5 - 1) ´ t = 6 l 6000 Þ t= = Å 3 3 Þ

t = 2000 Å = 0.2 mm

Hence correct option is (a). 12. Net shift = (m 1 - 1) t - (m 2 - 1) t = (m 1 - m 2 ) t = (1.52 - 1.40) ´ t Net shift = 0.12 ´ 10.4 mm

= 1248 ´ 10-9 m = 1248 nm Net shift = nl where n is + ve integer nl = 1248

Þ

Now, for 416 = l, n = 3 For 624 = l , n = 2 Hence correct option is (c). 13. w =

6300 Å ´ 1.33 m lD = d´ n 1 mm ´ 1.33 =

63 ´ 10-8 ´ 1.33 m 10-3 ´ 1.33

= 0.63 mm Hence correct option is (a). 7 Dl 3 Dl æ Dl ö 14. Dx = = 4ç ÷ nd nd è nd ø Dx = 4 ´ 0.63 mm = 2.52 mm Correct option is (a). 15. Dx = 2 of fridge width 2 ´ 0.63 mm 2 2 ´ 0.63 t= = 1.57 mm 0.53

(m - 1) t =

Hence correct option is (b). 16. Since on introducing thin glass sheet fringe width does not change hence fringe width = 0.63 mm. The correct option is (a).

57 ¢

More than one options correct

1. At centre path difference between all colours is zero hence cnetre is white since violet has least wavelength hence next to central will be violet and since intensity is different for different for colours hence there will be not a completely dark fringe Hence correct options, are (b), (c) and (d). 2. Correct options are (a), (c) and (d). q 3. Q I = 4 I 0 cos2 at centre q = 0 Þ I = 4 I 0 2 and at distance 4 mm above o is again maxima hence its intensity is also 4 I 0 .

4. The correct options, are (a), (c) and (d) Since phase difference = constant Light should be monochromatic. 5. Since in this case fringe pattern shift upward hence the correct options are (a), (b) and (c). 6. Q n1l1 = n2 l2 for maxima Using this option (a) is satisfied and (2n1 - 1) l1 = (2n2 - 1) l2 for minima Using this 3rd option is satified Hence correct options are (a) and (c).

Hence correct options are (a) and (c). ¢

Match the Columns

1. Q I = 4 I 0 cos2

q 2

(a) ® q; (b) ® p; (c) ® r; (d) ® s

if

q = 60°

I = 3I0

if

q = 90°

I = 2I 0

if

q = 0°

I = 4I0

if

q = 120°

I = I0

Hence correct match is (a) ® s; ( b) ® q; (c) ® p; (d)® r l 2. Dx = ´ Df 2p 2p l Þ Df = Dx if Dx = l 3 2p Þ Df = = 120° 3 l if Dx = 6 l Df = 60° if Dx = , Df = 90° 4 Dq ö Using I = 4 I 0 cos2 æç ÷ è 2 ø The correct match are

3. Distance between third order maxima and 3 Dl central maxima = = 3w d Distance between 3rd order minima and central 1 Dl Dl Maxima = æç 3 - ö÷ = 2.5 = 2.5 w 2ø d d è Distance between first minima and forth order 4 Dl Dl maxima = = = 2.5 w d 2d Distance between first minima and forth order 4 Dl Dl maxima = = 3.5 w d 2d Distance between 2nd order maxima and fifth order minima = 4.5 w - 2w = 2.5 w hence correct match is (a) ® q; (b) ® p; (c) ® r; (d) ® p

58 4. Since fringe shift in the division of sheet placed soure hence Similarly for other the correct match (b) ® r, s; (c) ® p; (d) ® p lD 5. When y = there will a dark fringe at 0. 2d hence (a) ® q lD æ lD ö when y = =3ç ÷ 6d è 2d ø The intensity becomes 3I (b) ® p

(c)® s

lD 3d

Intensity = I

(d) ® r

(a) ® p

when y =

when y =

lD 4d

Intensity = 2I

6. When a thin plate (transparent) is placed in front of S1 zero order fringe shift above from O hence (a) ® r When S1 is closed interference disappear and uniform illuminance is obtained on screen hence (b) ® p, q Similarly (c) ® r, s and When s is removed and two real sources s1 and s2 emitting light of same wavelength are placed interference disappear. Since sources become non-cohrrent hence (d) ® p, q

29. Modern Physics I Introductory Exercise 29.1 1. The positron has same mass m as the electron. The reduced mass of electron positron atom is m´ m 1 m= = m m+m 2 4

RH =

me 8e20 ch3

lP RH Þ = =2 lH RP Þ l p = 2lH = 2 ´ 6563 Å = 13126 Å

2.

Þ

1 1 = × z2 Þ lHe lH

Þ

lHe = 164 nm

ö × z2 ÷ ø lHe =

1 1 1 =Ré - ù l ëê 4 9 ûú 36 36 Þ l= = 5R 5 ´ 1.097 ´ 107 Þ l = 656 nm

Tn =

e n2 h2 ´ 2e 0 nh 2prn = 2p 0 un pme2 ´ e2

lH 6563 Å = 22 22

4 e 0 n 3h 3 me4 1 me4 = 2 3 3 Tn 4 e 0 n h

Þ r1 =

me4 4 e20 h3

Þ n1 =

9.1 ´ 10-31 ´ (16 . ´ 10-19 )4 4 ´ (8.85 ´ 10-12 )2 ´ (6.6 ´ 10-34 ) 3

Þ n1 = 6.58 ´ 1015 Hz n2 =

1 1 1 = R é 2 - 2 ù for largest wavelength n = 3 l n ûú ëê2 Þ

e2 2e 0 nh

rn =

1 1 1 = RP æç 2 - 2 ö÷ lP 3 ø è2

e 0 n2 h2 pme2

un =

=

R Þ RP = H 2 1 1 1 = RH æç 2 - 2 ö÷ lH 3 ø è2

= 1.31 mm 1 1 1 = RH æç 2 - 2 lHe 3 è2

3. For H-atom rn =

n1 n1 6.58 ´ 1015 = = 8 23 8

= 0.823 ´ 1015 Hz 1 1 1 (b) =Ré 2 - 2ù ê l 2 úû ë1 c 3 Þ n = = 3 ´ 108 ´ R é ù ê l ë 4 úû Þn=

9 ´ 108 ´ 1.097 ´ 107 4

= 2.46 ´ 1015 Hz (c) Number of revolutions = v2 ´ T = 0.823 ´ 1015 ´ 1 ´ 10-8 = 8.23 ´ 106 revolution

60 4. Reduce mass mm m p 207 m ´ 1836 m = = = 186 m mm + m p (207 m + 1836 m ) æ h2 r1 = 4 pe 0 çç 2 2 è 4 p me

æ ö h2 ÷÷ = 4 pe 0 ç 2 ç 4 p (186m) e2 ø è

ö ÷ ÷ ø

Putting the value we get -13

r1 = 2.55 ´ 10

36 = 653 nm 5 ´ 1.097 ´ 107

7. n K = n K a + n L a Þ lL a = b

m

Ionization energy = - E1 = 2.81 keV

6.6 ´ 10-34 (b) l = = 7.3 ´ 10-11m 9.1 ´ 10-31 ´ 107 6. (a) After absorbing 12.3 eV the atom excited to n = 3 state

Þ

b

hc DE

8. l =

E3 E2 = – 2870 eV Ka

Kb E1

lK a =

6.6 ´ 10-34 ´ 3 ´ 108 ( E1 - 2870) ´ 1.6 ´ 10-19

Þ 0.71 ´ 10-9 =

6.6 ´ 10-34 ´ 3 ´ 108 ( E1 - 2870) ´ 1.6 ´ 10-19

Þ

n=2

Solving this we get

n=1

E1 = - 4613 eV

1 1 = R é1 - 2 ù êë lL n úû 1 1 8R = R é1 - ù = êë lL 9 úû 9 1

9 9 = = 102 nm 8 R 8 ´ 1.097 ´ 107

lK

B

2

4 4 = = 122 nm 3 R 3 ´ 1.097 ´ 107

1 1 1 5R =Ré 2 - 2ù = êë2 lB 3 úû 36 Þ lB =

36 5R

=

6.6 ´ 10-34 ´ 3 ´ 108 = 0.63 ( 4613 - E3) ´ 1.6 ´ 10-19

Solving this we get E3 = - 2650 eV 4E 9. n 31 = =f h 5E

1 1 = R é1 - ù ê lL 4 úû ë 2 Þ lL =

b

1 1 1 l = 5.59 nm = lL a lK lK a L a

n=3

1

b

lK a - lK

0.71 ´ 0.63 c c c Þ lL a = = + 0.71 - 0.63 lK lK a lL a

6.6 ´ 10-34 h = = 4.8 ´ 10-34 m mv 46 ´ 10-3 ´ 30

Þ lL =

lK a ´ lK

b

-me4 E1 = 2 2 = - 2810 eV 8 e0 h

5. (a) l =

Þ lB =

3 4E 1

2 E

n21 =

3E 3 æ 4E ö 3 f = ç ÷= h 4è h ø 4

n 32 =

E f = h 4

…(i)

61

Introductory Exercise 29.2 1. eV0 =

hc -W l

Þ eV0 =

4. K max =

6.6 ´ 10-34 ´ 3 ´ 108 - 4.3 eV 2 ´ 10-7 ´ 1.6 ´ 10-19

é6.62 ´ 10-34 ´ 3 ´ 108 ù =ê - 3 ú eV -7 -19 ë 2 ´ 10 ´ 1.6 ´ 10 û

Þ eV0 = 6.2eV - 4.3 eV = 1.9 eV

= [6.20 - 3 ] eV = 3.20 eV

Þ V0 = 1.9 volt

The minimum kinetic energy = 0. -3

2. P = 1.5mW = 1.5 ´ 10

W

5.

K max = h [ f - f0 ]

Energy of each photon -34

=

6.62 ´ 10 ´ 3 ´ 10 4 ´ 10-7

Number of photons incident per second 1.5 ´ 10-3 P = Energy of each photon 4.96 ´ 10-19 ~ - 3 ´ 1015 The number of photoelectrons produce = 0.1% ´ 3 ´ 1015 = 3 ´ 1012 Current i = 3 ´ 1012 ´ 1.6 ´ 10-19 A -7

= 4.8 ´ 10

A = 0.48 mA

3. K max = hf - W = hf - hf0

…(i)

1.2 eV = h [ f - f0 ]

8

= 4.96 ´ 10-19 J

=

hc -W l

4.2 eV = h [1.5 f - f0 ]

…(ii)

Dividing Eq. (i) and Eq. (ii) f - f0 124 = 42 1.5 f - f0 Þ Þ Þ

3 f - 2 f0 = 7 f - 7 f0 5 f0 = 4 f 4 f0 = f = 0.8 f 5

éf ù Þ 1.2 ´ 1.6 ´ 10-19 = 6.62 ´ 10-34 ê 0 - f0 ú ë0.8 û Þ

1.2 ´ 1.6 ´ 10-19 2 f0 f0 = = 8 4 6.62 ´ 10-34

Þ

f0 = 1.16 ´ 1015 Hz

Þ K max µ ( f - f0 )

Subjective Questions (Level I) 1. Here l = 280 nm = 28 ´ 10-8 m E=

hc 6.6 ´ 10-34 ´ 3 ´ 108 = J l 28 ´ 10-8

E=

19.8 ´ 10-34 ´ 1016 198 ´ 10-19 J= J 28 28 -19

E=

198 ´ 10 198 ~ eV = - 4.6 eV -19 28 ´ 1.6 28 ´ 1.6 ´ 10

We have E = mc2 Þ m =

198 ´ 10-19 E = c2 28 ´ 9 ´ 1016

Þ m = 8.2 ´ 10-36 kg Momentum p = mc = 8.2 ´ 10-36 ´ 3 ´ 108 = 2.46 ´ 10-27 kg-m/s 2. Intensity of light at a distance 2 m 1 1 From the source = = W /m 2 2 16p 4 p ´ (2)

62 5. (a) E = 2.45 MeV = 2.45 ´ 1.6 ´ 10-19 ´ 106 J

Let plate area is A Energy incident on unit time is 1 E1 = Aw 16p Energy of each photon -34

=

8

6.6 ´ 10 ´ 3 ´ 10 4.8 ´ 10-7

Number of photons striking per unit area 1 ´ A ´ 4.8 ´ 10-7 16 p n= A ´ 6.6 ´ 10- 34 ´ 3 ´ 108 ´ A = 4.82 ´ 1016 per m 2 s

E = hn Þ n =

Þ n = 5.92 ´ 1020 Hz (b) We have c = nl Þ l = Þl= 6. We have

Þ E = 2.47 ´ 10

J E in joule Energy in eV = 1.6 ´ 10-19 2.47 ´ 10-19 = 1.6 ´ 10-19 Energy (in eV) = 1.54 eV

(b) Wavelength l=

6.6 ´ 10-34 h = = 804 nm p 8.24 ´ 10-28

This wavelength in Infrared region. 4. We have c = f l Þ f =

c 3 ´ 108 m/s = l 6 ´ 10-7 m

f = 5 ´ 1014 Hz E We have p = Þ E = pt power per sec = t energy

Þ

Þ

P=E

Þ

75 = ( h ´ v) ´ n 75 = -34 6.6 ´ 10 ´ 5 ´ 1014

Þ Þ

n = 2.3 ´ 1020 photons/sec

3 ´ 108 = 5.06 ´ 10-13 m 5.92 ´ 1020 p = 2mK p1 = 2mK 1 and p2 = 2mK 2

Þ

p1 = p2

K1 K2

Þ

1 = 2

K1 K2

(a) Energy of photon E = pc -19

c n

Þ

3. Here p = 8.24 ´ 10-28 kg-m/s

Þ E = 8.24 ´ 10-28 ´ 3 ´ 108

E 3.92 ´ 10-13 = h 6.6 ´ 10-34

[Q p2 = 2 p1 ]

K2 = 4 K1

Þ

(b) E1 = p1c and E2 = p2 c Þ E2 = 2 p1c Þ E2 = 2E1 7. (a) Since power = energy per unit line let n be the number of photons nc Þ P = nE Þ 10 = n ´ l 10 ´ l 10 ´ 500 ´ 10-9 Þn= = hc 6.6 ´ 10-34 ´ 3 ´ 108 Þ n = 2.52 ´ 1019 (b) Force exerted on that surface P 10 F= = = 3.33 ´ 10-8 N c 3 ´ 108 8. Absorbing (power) light = 70% of incident light Þ Pa = 70% of 10 W = 7 W Refractive power = 30% of 10 W Þ PR = 3 W

Pa 2 PR + c c 7 + 2´ 3 13 = = c 3 ´ 108

The force exerted =

= 4.3 ´ 10-8 N

63 9. Force = rate of change of momentum p

13. (a) l =

h h 6.6 ´ 10-34 Þ p= = l 2.8 ´ 10-10 p

p cos 60°

Þ p = 2.37 ´ 10-24 kg-m/s p sin 60

(b) Q p2 = 2me K Þ K =

60 p cos 60

ÞK = p sin 60°

p2 2me

(2.37 ´ 10-24 )2 2 ´ 9.1 ´ 10-31

Þ K = 3.07 ´ 10-18 J

Dp here Dt = 1 s F= Dt

K (in eV ) =

K in J 3.07 ´ 10-18 = 1.6 ´ 10-19 1.6 ´ 10-19

F = 2 p cos 60° F=p nh F= l

Þ

14. Here T = 273 ° + 20° = 293 K vrms =

where n is number of photons striking per second

Þl=

1 ´ 1019 ´ 6.63 ´ 10-34 Þ F= = 10-8 N 663 ´ 10-9 10. Here output energy = 60 W/s Pressure p =

2 ´ 60 = 4 ´ 10-7 N 3 ´ 108

K = 19.2 eV

Þ

=

3 ´ 1836 ´ 9.1 ´ 10-31 ´ 8.31 ´ 293

Þ l = 1.04 Å 15. For hydrogen like atom E = - K Here E = - 3.4 eV K = 3.4 eV = 3.4 ´ 1.6 ´ 10-19 J h h l= = p 2me K

Þ

by de-Broglie hypothesis wavelength l=

6.62 ´ 10-34 h = mv 5 ´ 10-3 ´ 340

l = 3.9 ´ 10-34 m

Þ

Since l is too small. No wave like property is exhibit. 6.6 ´ 10-34 h 12.(a) le = = mev 9.1 ´ 10-31 ´ 4.7 ´ 106 -10

= 1.55 ´ 10 (b) l p =

m

6.6 ´ 10-34 1836 ´ 9.1 ´ 10-31 ´ 4.7 ´ 10-6 = 8.44 ´ 10-14 m

h 3 MRT 6.6 ´ 10-34

de-Broglie wavelength 11. Here m = 5 g Þ m = 5 ´ 10-3 kg, v = 340 m/s

3 RT h Þl= M Mvrms

=

6.6 ´ 10-34 2 ´ 9.1 ´ 10-31 ´ 3.4 ´ 1.6 ´ 10-19

Þ l = 6.663 Å 16. In Bohr model the velocity of electron in nth orbit is given by Un =

e2 2e 0 nh

Putting the values of e, e 0 , h and n = 1, we get U1 = 2.19 ´ 107 m/s and U4 =

2.19 ´ 106 m/s 4

64 Þ l1 =

h h h and l4 = = = 4 l1 mev1 meu4 m u1 e 4

Þ

l1 =

6.6 ´ 10-34 9.1 ´ 10-31 ´ 2.19 ´ 106

= 3.32 ´ 10-10 m l4 = 4 l1 = 4 ´ 3.332 ´ 10-10 m

Þ

= 1.33 ´ 10-9 m The radius of first Bohr orbit r1 = 0.529 ´ 10-10 The radius of fourth Bohr orbit r4 = 16 ´ 0.529 ´ 10-10 2pr1 = 2 ´ 3.14 ´ 0529 . ´ 10-10

Þ

» 3.32 ´ 10-10 m = l1

= 13.28 ´ 10-10 m

Bohr Atomic Model and Emission Spectrum 17. For hydrogen like atom we can write en =

- z2 (13.6 eV) n2

For lithium atom z = 3 we get -9 -122.4 En = 2 (13.6 eV) = eV n n2 The ground state energy is for n = 1 -122.4 Þ E1 = eV = - 122.4 eV 12 Ionization potential = - E1 = 122.4 eV 18. For hydrogen atom we can write (a) E = - K Þ K = 3.4 eV (b) PE = - 2 K = - 2 ´ 3.4 = - 6.8 eV Since potential energy depends upon refrence hence it will changed. 19. Binding energy of an electron in He-atom is E0 = 24.6 eV. ie, the energy required to remove one electron from He-atom = 24.6 eV Now, He-atom becomes He + and energy of He + ion is given by En =

- z2 (13.6) for He + z = 2, we get n2

E1 = - 4 ´ 13.6 = - 54.4 eV. Hence energy required to remove this electron = 54.4 eV, thus total energy = 24.6 + 54.4 = 79 eV

20. For hydrogen atom En =

-13.6 eV n2

Putting n = 3, we get -13.6 eV E3 = = - 1.51 eV 9 Hence hydrogen atom is in third excited state the angular momentum L=

nh 3 h 3 ´ 6.62 ´ 10-34 = = 2p 2p 2 ´ 3.14

Þ L = 3.16 ´ 10-34 kg-m 2 /s hc 6.6 ´ 10-34 ´ 3 ´ 108 = DE DE 6.6 ´ 10-34 ´ 3 ´ 108 (in Joule) Þ DE = 1023 ´ 10-10

21. We have l =

Þ DE =

6.6 ´ 10-34 ´ 3 ´ 108 (in eV) 1023 ´ 10-10 ´ 1.6 ´ 10-19

~ - 12.1 eV Þ En - E1 = 12.1 eV But E1 = - 13.6 eV Þ En ~ - - 1.51 eV -13.6 13.6 For H-atom En = Þ - 151 . =- 2 2 n n Þ n =3 Hence atom goes to 3rd excited state. The possible transition are (3 ® 2, 3 ® 1, and 2 ® 1) ie, 3 transitions are possible and the largest wavelength = 1023 Å (From 3 ® 1)

65 Let for ten transitions quantum numbers of energy levels are n, n + 1, n + 2, n + 3 and n+4

22. For hydrogen like atom 2

En =

-z (13.6) eV n2

– 0.544 eV

For Li + + z = 3 -122.4 Þ En = eV Þ E1 = - 122.4 eV n2 -122.4 E3 = eV = - 13.6 eV 9

n+1 n+2 n+3

DE = E3 - E1 = (122.4 - 13.6) eV = 108.8 eV -34

l=

- z (13.6) eV = - 0.85 eV n2 - z2 (13.6) eV = - 0.544 eV ( n + 4)2

Þ

83.7 ´ 10-19 J = = 52.3 eV …(i) 1.6 ´ 10-19 J/eV

Þ Þ

For He + z = 2

Putting this value of n in Eq. (i)

54.4 En = 2 eV n Þ

E1 = - 54.4 eV

Þ

1 En - E1 = 54.4 æç 1 - 2 ö÷ eV n ø è

From Eqs. (i) and (ii), we get 1 52.3 = 54.4 æç 1 - 2 ö÷ n è ø 1 Þ 1 - 2 = 0.96 n n =5

24. For hydrogen like atom En =

…(ii)

( n + 4)2 0.85 = = 1.5625 2 0.544 n n+4 = 1.25 n 4 1 + = 1.25 n 4 = 0.25 n 4 n= = 16 0.25

Þ

- z2 (13.6) eV n2

…(i)

Dividing Eq. (i) and Eq. (ii)

En - E1 = 83.7 ´ 10-19 J

For He + atom En =

Þ

2

hc 6.6 ´ 10 ´ 3 ´ 10 = = 113.74 Å E 108.8 ´ 1.6 ´ 10-19

1 ù 663 . ´ 10-34 ´ 3 ´ 108 é 1 + 10-10 ëê1085 304 ûú

n+4

– 0.85 eV

8

23. The excited state energy He + atom will be equal to the sum of energies of the photons having wavelength 108.5 nm and 30.4 nm. hc hc Þ En - E1 = + l1 l2

Þ

n

- z2 (13.6) eV n2

…(ii)

- z2 (13.6) = - 0.85 (16)2 256 ´ 0.85 Þ z2 = 13.6 z2 = 16 Þ z = 4

Þ

Hence atom no. of atom is = 4 hc We know that DE = l hc for smallest wavelength DE is Þl= DE maximum l=

6.6 ´ 10-34 ´ 3 ´ 108 [ - 0.544 - ( - 0.85)] ´ 1.6 ´ 10-19

l = 40441 Å

66 25. Here E1 = - 15.6 eV (a) Hence ionization potential = - E1 = 15.6 eV hc (b) We have l = for short wavelength DE DE is maximum Þl=

6.6 ´ 10-34 ´ 3 ´ 108 ~ - 2335 Å 0 - ( - 5.3) ´ 1.6 ´ 10-19

(c) Excitation potential for n = 3 state is = E3 - E1 = - 3.08 + 15.6 = 12.52 V (d) From n = 3 to n = 1 DE = E3 - E1 = - 3.08 + 15.6 = 12.52 eV hc We know l = DE Þ

12.52 ´ 1.6 ´ 10-19 1 DE = = l hc 6.6 ´ 10-34 ´ 3 ´ 108 = 1.01 ´ 107 (m

v2 = mw2 r2

Þ

-1

)

But by Bohr’s postulate mvr = Þ

m2v2 r2 =

n2 h2 4 p2

Þ

m 3w2 r4 =

n2 h2 4 p2

Þ

r4 =

Þ 1 28. = R ( z - 1)2 lK a

Energy of this photon 12375 (eV) = = 1.44 eV 8600 hence internal energy of atom after absorbing this photon is given by Ei = E1 + 1.44 eV = - 6.52 + 1.44 = - 5.08 eV 12375 (eV) (b) l2 = = 2.95 eV 4200 hence internal energy of the atom after emission of this photon is given by Ei = E1 - 2.95 eV = ( - 2.68 - 2.95) eV Ei = - 5.63 eV 1 dU 27. Heve U = mw2 r2 Þ F = = m2 w2 r 2 dr Þ

2

But

mv = m2 w2 r r

rµ n é1 - 1 ù = 1 êë 22 úû l0

b

lK

b

lK a

Þ

=

lK = b

26. (a) E1 = - 6.52 eV l = 860 nm = 8600 Å

n2 h2 4 p2 m 3w2

1 = R ( z - 1)2 lK Þ

29. l0 =

nh 2p

é1 - 1 ù êë 32 úû

3 ´ 9 27 = 4 ´ 8 32 27 27 lK a = l0 32 32

hc eV

Þ l0 1 =

6.6 ´ 10-34 ´ 3 ´ 108 1.6 ´ 10-19 ´ 25 ´ 103

Þ l01 = 49.5 pm Þ l0 2 = 2l01 = 2 ´ 49.5 pm = 99 pm [1 pm = 10-12 m ] 30. fKa = (2.48 ´ 1015 ) Hz ( z - 1)2 Þ

3 ´ 108 = 2.48 ´ 1015 (2 - 1)2 lK a

Þ

3 ´ 108 = ( z - 1)2 0.76 ´ 10-10 ´ 2.58 ´ 1015

( z - 1) ~ - 40 Þ z = 41 hc 31. li = Dl = 26 pm when V f = 1.5 V eV Þ

l f = ( li - 26) pm hc 1 hc 12 1 lf = = = = li e ´ 1.5 V 1.5 eV 1.5 1.5

67 2 li 3

1 a2 = l26 c

Þ

( li - 26) =

Þ

3 li - 26 ´ 3 = 2li

Þ

li = 78 pm

Þ 78 ´ 10-12 = Þ V =

2

5 f æç 26 - ö÷ 1 3 è ø = 2 l26 5 887 pm ´ æç 13 - ö÷ 3ø è

6.6 ´ 10-34 ´ 3 ´ 108 1.6 ´ 10-19 ´ V

2

6.6 ´ 3 ´ 10-26 1.6 ´ 78 ´ 10-12 ´ 10-19

Þ

l26

Þ V = 15865 volt 32.

2

æ 26 - 5 ö ç ÷ 3ø è

5 887 pm ´ æç 13 - ö÷ 3 è ø = 2 æ 26 - 5 ö ç ÷ 3ø è

V = a ( z - b) Þ Þ

= 887 pm ´

c = a2 ( z - b)2 l

~ - 198 pm

1 a2 = ( z - b)2 l c

33.

Þ

1 a2 = (13 - b)2 887 pm c

…(i)

and

1 a2 = (30 - b)2 146 pm z

…(ii)

Dividing Eq. (i) and Eq. (ii)

Þ Þ

146 é(13 - b) ù =ê ú 887 ë (30 - b) û 30 - b 2.5 = 13 - b

342 (73)2

f =

3 RC ( z - 1) 4

4.2 ´ 1018 =

3 ´ 1.1 ´ 107 ´ 3 ´ 108 ( z - 1)2 4

4.2 ´ 1018 ´ 4 = ( z - 1)2 9 ´ 1.1 ´ 1015

Þ

Þ ( z - 1) = 41 Þ z = 42

2

34. P = Vi = 40 kW ´ 10 mA = 400 W % of P =

400 ´ 1 = 4W 100

(a) Total power of X-rays = 4 W

32.5 - 2.5b = 30 - b

(b) Heat produced per second

2.5 = 1.5b 5 b= 3

= 400 - 4 = 396 J/s

Photoelectric effect 35. Einstein photo electric equation is K max = hn - W hc Þ eV0 = - W Q K max = eV0 l

Þ 10.4 eV =

12375 l( Å)

- 1.7 eV

12375 = 1022 Å 12.1 hc 12375 For H-atom l = Þ DE = = 12.1 eV DE 1022

Þ l( Å) =

68 This difference equal to n = 3 ® n = 1 transition.

Þ h=

36. K max = hn - W Þ K max =

39. Here

6.6 ´ 10-34 ´ 1.5 ´ 1015 - 3.7 1.6 ´ 10-19

Þ K max = 6.18 - 3.7 = 2.48 eV

K max = hn - W Kmax (eV) C

6

2 A 10

20

1 hc m [v2 (max) ]2 = -W 2 l2

D 30

9 hc hc = 8W l2 l1

Þ

9 1 ù hc é = 8W êë6000 3000 úû

Þ

6.6 ´ 10-34 ´ 3 ´ 108 ´ 7 = 8W 6000 ´ 10-10 ´ 1.6 ´ 10-19

Þ

W = 1.81 eV

f = 1 ´ 1014 Hz

-181 . ´ 16 . ´ 10-19 14

(a) n o = threshold frequency q = 10 ´ 10 Hz (b) W = 4 eV (c) h = slope of the graph 8 eV CD = = AD 20 ´ 1014 Hz

…(ii)

Putting the value of W in Eq. (i) 1 hc m(u1 max )2 = 2 3000 ´ 10-10

–4

= 1015 Hz

…(i)

Þ

4

0

hc - W, l

Dividing Eq. (i) and Eq. (ii), we get hc hc 2 -W -W æ v1 max ö l ç ÷ = l1 Þ (3)2 = 1 ç v2 max ÷ hc hc è ø -W -W l2 l2

38. Comparing the given graph with

–2

3 1

Using Einstein equation, K max =

and

12375 l= = 2260 Å 5.475

8

u2 (max)

=

where m is the mass of photoelectron 1 hc Þ m[v1(max) ]2 = -W 2 l1

K max = eV0 = 3 eV hc K max = -W l 12375 3= - 2.475 l(in Å)

Þ

v1(max)

we get 1 hc mv2max = -W 2 l

37. Here work function 12375 W(in eV) = = 2.475 eV 5000 Å

Þ

8 ´ 1.6 ´ 10-19 = 6.4 ´ 10-34 J-s 2 ´ 1015

= 34

662 . ´ 10 ´ 3 ´ 108 1 m(u1 max )2 = 2 3 ´ 107 -2896 . ´ 10-19 1 m(u1 max )2 = 662 . ´ 10-19 - 2896 . ´ 10-19 2 1 m(u1 max )2 = 3.724 ´ 10-19 2

69 Þ (u1 max )2 =

3.724 ´ 10-19 ´ 2 9.1 ´ 10-31

-1.81 ´ 16 . ´ 10-19 Q

me = 9.1 ´ 10-31

Þ u1 max = 9 ´ 105 m/s 1 and v2 max = v1 max = 3 ´ 105 m/s 3 40. Here intensity I = 2 W /m 2 and -4

Area A = 1 ´ 10 m

vmax = r=

42. The given equation is E = (100 v / m ) [sin ((5 ´ 1015 ) t

2

+ sin (8 ´ 1015 ) t ] Light consist of two different frequencies. Maximum frequency 8 ´ 1015 = = 1.27 ´ 1015 Hz 2p

E = IA = 2 ´ 10-4 W 2 ´ 10-4 eV = 2 ´ 10-4 J/s = 1.6 ´ 10-19 s

For maximum KE we will use Einstein’s equation

2 ´ 1015 eV/s 1.6

(KE) max = hn - W

Energy of each photon = 10.6 eV

=

Number of photons incident on surface =

2 ´ 1015 1.6 ´ 106 .

Number of photoelectrons emitted

43. Here E = E0 sin 1.57 ´ 107 ( x - ct) frequency of the wave n=

= 6.25 ´ 1011 per second Maximum KE = (10.6 - 5.6) eV = 5 eV hc 41. K max = -W l

K max

Þ ¢

6.6 ´ 3 ´ 10-18 1 mev2max = - 2 ´ 1.6 ´ 10-19 2 18 1 mev2max = 11 ´ 10-19 - 3.2 ´ 10-19 2 = 7.2 ´ 10-19 J

1.57 ´ 3 ´ 1015 = 0.75 ´ 1015 2 ´ 3.14

We have æ 6.6 ´ 10-34 ´ 0.75 ´ 1015 ö eV0 = çç - 1.9 ÷÷ eV -19 1.8 ´ 10 è ø

Minimum KE = 0

6.6 ´ 1034 ´ 3 ´ 108 = - 2 ´ 1.6 ´ 10-19 180 ´ 10-9

1.27 ´ 1015 ´ 6.62 ´ 10-34 -2 1.6 ´ 10-19

Þ (KE) max = 3.27 eV

0.53 2 ´ 1015 = ´ 100 1.6 ´ 106 .

Þ

mvmax 9.1 ´ 10-31 ´ 1.25 ´ 106 = eB 1.6 ´ 10-19 ´ 5 ´ 10-5

r = 0.148 m

Energy incident per unit time on the metal surface

=

2 ´ 7.2 ´ 10-19 = 1.25 ´ 106 m/s 9.1 ´ 10-31

V0 = 1.2 V

Objective Questions (Level-1)

1. Einstein photo electric equation is K max = hn - W its slope = h = planck constant which is same for all metals and independent of intensity of radiation. Hence correct option is (d). 2. Since current is directly proportional to intensity therefore as current is increased

70 intensity is increased since lmin µ

1 , if V is V

decreased lmin is increased.

Þ

h l2 mv2 v3 = = n l3 v2 mv3

Þ

l2 2 = l3 3

Hence correct option is (c). 3. For hydrogen atom (Bohr’s model) nth e2 orbital speed vn = 2e 0 nh For first orbit n = 1

8. For hydrogen like atom

e2 v1 = 2e 0 ´ h

Þ

=

En =

(1.6 ´ 10-19 )2 2 ´ 8.85 ´ 10-12 ´ 6.62 ´ 10-34

1 ö c 8 v~ - æç ÷ ´ 3 ´ 10 = 137 è 137 ø

Þ

Hence correct option is (c). 4.

86

A

22

¾® X 3a 80

210

4b

210

¾® 84 B

Hence correct option is (b) 12375 12375 ~ 5. lmin (in Å) = = - 0.62 Å V (in volt) 20 ´ 1000 Hence correct option is (c). 1 6. We have mev2max = eV 2 Þ

vmax

2eV = m =

Þ

Hence correct option is (a).

For ground state n = 1 Þ

E1 = - z2 ´ 13.6 eV

But

E1 = - 122.4 eV

Þ -122.4 eV = - z2 ´ 13.6 eV Þ

z2 = 9

Þ

z =3

Hence it is Li2 + The correct option is (c). hc Dlmin DV 9. lmin = Þ ´ 100 = ´ 100 eV lmin V Percentage change in lmin = - 2% Hence lmin is decreased by 2% correct option is (c)

2 ´ 1.6 ´ 10-19 ´ 18 ´ 1000 9.1 ´ 10-31

10. En =

Hence correct option is (a). e2 2e 0 nh

Þ

e2 e2 v2 = , v3 = 2e 0 ´ 2h 2e 0 ´ 3 h

Þ

v2 3 = v3 2

Let l2 and l3 are the de-Broglie wavelengths

- z2 (13.6 eV) for first excited state n = 2 n2

Þ

E2 =

- z2 (13.6) eV 4

Þ

-13.6 eV =

- z2 ´ 13.6 eV 4

vmax ~ - 8 ´ 107 m/s

7. For hydrogen atom vn =

- z2 (13.6 eV) n2

Þ

z =2

Hence it is He + Correct option is (a). 12375 11. lmin (in Å) = V (in volt) 12375 = 12.375 ´ 103 V 1

Þ

V =

Þ

V = 12.4 eV

71 Hence correct option is (c). h 12. We have l = p

eV0 = hc ´ =

-34

Þ

p=

h 6.62 ´ 10 = l 0.5 ´ 10-10

V0 =

Þ l = 13.26 ´ 10-24 kg-m/s Hence correct option is (c). 13. W = Þ

Þ

l0 = 7750 Å

Hence correct option is (a). Balmer

series

is

Þ

30 15 = 16 8 15 V0 = V 8

4E =

3 hc -W l

z =2

Þ Þ

= 54.4 eV Hence correct option is (a).

…(ii)

hc = 3W l hc W = 3l

Hence correct option is (b).

15. If V1 = 0 then total energy = KE

18. By Moseley’s law f = a ( z - b) for K a line b = 1

KE = 13.6 eV

f = a ( z - 1)

and the energy difference between two states = 10.2 eV

Þ

Hence total energy in this state

and

f ¢ = a (51 - 1) = 9 ´ 50

Þ

f 3 = f¢ 5

= 13.6 + 10.2 = 23.8 eV The correct option is (c). hc 16. eV0 = -W 3300 hc 2eV0 = 2200

…(i)

From Eqs. (i) and (ii) æ hc ö 3 hc 4ç -W ÷ -W l è l ø

Binding energy = (13.6) ´ z2eV = 13.6 ´ 4 eV

Þ

6.6 ´ 10-34 ´ 3 ´ 108 3 ´ 2200 ´ 1.6 ´ 10-19 ´ 10-10

Hence the correct option is (c). hc 17. E = -W l hc 4E = -W l/3

For third Balmer line n = 5 1 1 1 Þ = z2 ´ 109 . ´ 107 é - 2 ù -10 ê 1085 ´ 10 ë 4 5 úû 100 ´ 1000 Þ z2 = =4 1085 ´ 1.097 ´ 21 Þ

hc 3 ´ 2200 ´ 10-10

=

66 . ´ 10-34 ´ 3 ´ 108 hc Þ l0 = l0 16 . ´ 10-19 ´ 16 .

14. For H-like atom 1 1 1 = z2 R é 2 - 2 ù ê l n úû ë2

3300 - 2200 3300 ´ 2200

Þ

f = a (31 - 1) = a ´ 30

f¢ =

…(i)

25 f 9

Hence correct option is (a). …(ii)

Substracted Eq. (i) from Eq. (ii), we get

19.

1 1 f = ( RC) ( z - 1) é - 2 ù ëê12 n ûú for K a , n = 2

…(i)

72 K b, n = 3

Þ

1 fa = RC ( z - 1) é1 - 2 ù = RC ( z - 1) ´ 2 ûú ëê

3 4

W =

hc 6l

Hence correct option is (a). 23. eV0 = h [2V0 - V0 ] = hV0

fb = RC ( z - 1) 1 fb fa

1 8 = RC ( z - 1) ´ 9 9

8´ 4 32 = = , 9´3 27

From Eqs. (i) and (ii) Hence correct option is (b). 24. For H-atom Lyman series is 1 1 1 =Ré 2 - 2ù l n ûú ëê1

æ9 - 4 ö 1 1 1 5R ÷R = = R é 2 - 2 ù = çç l 36 3 ûú è 9 ´ 4 ÷ø ëê2 36 Þ l= 5R Þ

Þ

For Li + + , n = 3 Þ n¢ = n ´ 32 = 9 v Hence correct option is (c). 25. Ground state energy of H-atom = - 13.6 eV

1.8 ´ 1011 ´ 10000 2

From Eqs. (i) and (ii) 5hc hc - 5W = -W 3l l 5hc hc Þ =4W 3l l hc (5 - 3) Þ =4W l 3 2hc Þ = 4W 3l

( -13.6) eV ´ z2 n2 -13.6 eV ´ 9 -13.6 eV = n2

For Li + + atom En = Þ

3 ´ 107 lD 1 = = lmin 3 ´ 108 10

Hence correct option is (c). hc 22. 5 eV0 = -W l hc eV0 = -W 3l

c 1 1 = RC é 2 - 2 ù l n ûú ëê1

1 1 n¢ = RC é - 2 ù ´ z2 ëê12 n ûú

lD 1 eV = lmin c 2m 1 = 3 ´ 108

v=

For H-like atom

Hence correct option is (c) h hc 21. lD = and lmin = eV 2 meV

Þ

eV = h [3 V0 - V0 ] = h ´ 2V0 = 2hV0 …(ii) eV = 2 ´ eV0 Þ V = 2V0

Correct option is (a). 1 1 1 20. = R é 2 - 2 ù here n = 3 ê l n úû ë2

Þ

…(i)

Þ

…(i) …(ii)

n =3

Hence correct option is (c). 1 26. mv12 = hn1 - W 2 1 mv22 = hn2 - W 2 1 Þ m [v12 - v22 ] = h [n1 - n2 ] 2 2h Þ v12 - v22 = [n1 - n2 ] m Hence correct option is (b). 27. For Lyman series 1 1 1 =Ré 2 - 2ù êë1 l n úû

…(i) …(ii)

73 For largest wavelength n = 2 1 1 1 =Ré 2 - 2ù ê l 2 úû ë1 1 1 Þ = R é1 - ù ê l 4 úû ë 4 Þ l= 3R

Hence correct option is (c). h h 31. We have, le = , la = 2me E1 2ma E2 lp = and

For He + atom 1 1 1 1 1 = R ( z)2 é 2 - 2 ù = 4 R é - 2 ù êë2 ú ê lHe n û ë 4 n úû 3R 1 1 = 4R é - 2 ù êë 4 n úû 4 3 1 1 = - 2 16 4 n 4 -3 1 1 3 = = 16 n2 4 16 1 1 = n2 16

Þ Þ Þ Þ

le = la = l p Q E1 > E3 > E2

Hence correct option is (a). 32. KE in ground state = 13.6 eV Total energy in n = ¥ = KE + Energy difference between n = ¥ to n = 1

= 27.2 eV Hence correct option is (b). 33. Here P = 1000 W, n = 880 kHz = 880 ´ 103Hz Let n is the number of photon p emitted per second P 1000 Þ n= = -34 hn 6.62 ´ 10 ´ 880 ´ 103

1 3R = ´ ( z - 1) l 4

3 ´ 1.0973 ´ 107 1 Þ = ´ (92 - 1) l 4 4 Þ l= = 0.15 Å 3 ´ 91 ´ 91 ´ 1.0973 ´ 107

= 1.7 ´ 1030 Correct option is (b). 34. Here l = 3000 Å = 3 ´ 10-7 m Energy of incident radiation E =

Hence correct option is (c). Þ

29. For K a , K b and La of X-rays nK = nK a + nL a b

Y2 = Y1 + Y3

Þ

Hence correct option is (b). h 30. We have l = 2mqV lp =

Þ

lp la

=

h 2m p eV

and la =

Q ma > m p > me

= 13.6 eV + 13.6 eV

Hence correct option is (b). 28. We have

2m p E3

Þ Total energy

n=4

Þ

h

hc joule l

E=

hc (in eV) l ´ 1.6 ´ 10-19

E=

6.62 ´ 3 ´ 10-26 = 4.125 eV 3 ´ 1.6 ´ 10-26

Q E < work function hence no emission of electrons it means sphere remain natured. h 2ma (2 e) V

2 ´ 4mp 2ma = = 8 =2 2 mp mp

Hence correct option is (c). -13.6 eV 35. Q En = for n = 5 n2 -13.6 eV E5 = = - 0.54 eV 52 Hence correct option is (a).

74 36. K max = E - W = (6.2 - 4.2) eV = 2 eV

é 6.62 ´ 10-34 ´ 3 ´ 108 ù 38. K max = ê - 1ú eV -10 -19 ë3000 ´ 10 ´ 1.6 ´ 10 û

Þ K max = 2 ´ 1.6 ´ 10-19 J = 3.2 ´ 10-19 J

K max = 3.14 eV = 3.14 ´ 1.6 ´ 10-19 J 1 Þ mev2max = 3.14 ´ 1.6 ´ 10-19 2

Hence correct option is (b). 37. l =

6.62 ´ 10-34 = 9.1 ´ 10-31 ´ v 5200 ´ 10-10

Þ

Hence correct option is (c).

6.625 ´ 10-27 ~ v= - 1400 m/s 5.2 ´ 9.1 ´ 10-31

Þ vmax =

3.14 ´ 3.2 ´ 10-19 ~ 6 - 10 m/s 9.1 ´ 10-31

Hence correct option is (d).

JEE Corner 1. Here both assertion and reason are true and reason explain correctly assertion. Correct option is (a). hc E 2. For photon E = and p = l c Þ If l is doubled, E and p are reduced to half. Hence assertion is true. Since speed of photon is always c. Hence reason is false. Hence correct option is (c). 3. If frequency is increased keeping intensity constant photoelectron emitted the plate reach other plate in less time hence saturation current can be increased. Reason can be true or not hence correct option is (a, b). 4. Here both assertion and reason is true and reason correctly explain assertion. Hence correct option is (a). 5. Here assertion is true since possible transition are 6 ® 3, 6 ® 4, 6 ® 5, 5 ® 3, 5 ® 4, and 4 ® 3. According to reason total n( n - 1) 6 ´ (3 - 1) transition has n = 3 Þ =6 2 2 it may explain or may not explain assertion Hence correct option is (a, b) 6. We have eV0 = h[n - n 0 ]

Þ

V0 =

h h n0 - n0 e e

…(i)

if n ® 2 n 0 n 0 does not become double hence assertion is false but reason is true. Hence correct option is (d). 7. Here both assertion and reason are true and reason may or may not explain assertion correct option is (a, b). 8. Here assertion and reason are both true. hc if V ® increases Q lmin = eV lmin ® decreases but reason is not correct explanation of assertion hence correct option is (b). -13.6 9. Q En = n2 Þ

E2 > E1

Hence assertion is true and E = - K = +

v 2

Þ v is more in n = 2 Here reason is also true but it is not correct explanation of assertion hence correct option is (b). 10. Here assertion is false but reason is true. Hence correct option is (a).

75 ¢

Objective Questions (Level 2)

1. Fa = Fc Þ and

GmM mv2 = r r2 nh mvr = 2p

3. For H-like atom En = …(i) Here E2 = …(ii)

From Eq. (ii) nh Þv= 2pmr

Þ Þ

-(13.6) z2 and E1 = - 13.6 z2 4 E2 - E1 = 40.8 eV 1 13.6 Z 2 ´ é1 - ù = 40.8 4 ûú ëê Z2 =

Putting this value in Eq. (i) GM h2 h2 = 2 2 2 r 4p m r

Þ

2

- GMm - GMm ´ 4 p m GM = 2r 2n2 h2

-2p2G2 M 2 m 3 for ground state n = 1 ÞE= n2 h2 2

Þ E=

2

2

- 2p G M m n2

3

Hence correct option is (b). e 2. We have m n = in An = ´ p rn2 Tn e ´ p ´ rn2 ´ un e un rn Þ mn = = 2prn 2 Þ mn = e ´ Þ and Þ

Z =2

Hence correct option is (a). e e ev in = = = n Tn 2prn 2prn un

Þ E = KE + PE Þ E=

408 . ´4 Þ Z2 = 4 13.6 ´ 3

Energy needed to remove the electron from ground state is - E1 = + (13.6) ´ Z 2 = + 13.6 ´ 4 = 54.4 eV

n2 h2 Þ r= 2 2 4p m GM 1 1 m ´ GM GMm KE = mv2 = = 2 2 r 2r -GMm PE = r 2

- (13.6)Z2eV n2

v1 r1 ´ n2 e v1r1 ´ = ´n n 2 2 ev r ´ 2 m2 = 1 1 2 e v1r1 m1 = ´1 2 m m1 = 2 2

Hence magnetic moment decreases two times correct option is (b).

Q Þ Þ

1 1 and rn µ 2 n n 1 i 1 in µ 3 Þ 2 = 3 i1 2 n

un µ

i1 = 8 i2

Hence current increases 8 times correct option is (c). 5. Since five dark lines are possible hence atom is excited to n = 6 state. The number of transition in emission line n ( n - 1) = 2 6´ 5 Number of emission transition = = 15 2 Hence correct option is (c). 6. An = prn2 for hydrozen atom rn = kn2 where k is constant. Þ

An = pk2 n4

Þ

A1 = pk2 µ

An = n4 A1

76 1.5 ´ 1015 6.6 ´ 10-34 ´ 2p 1.6 ´ 10-19

æA ö Taking log both sides log çç n ÷÷ = 4 log n è A1 ø

=

Hence it is a straight line with slope = 4

= 0.98 eV

Correct option is (b) 1 7. For hydrogen atom in µ 3 and n in 1 Bn µ µ Bn = k 5 rn n Þ Þ Þ

Since work function = 2 eV > maximum energy hence no emission of electrons. Thus correct option is (d). 2

[Q rn ´ n ]

k and B1 = k 25 B2 1 1 = = B1 25 32

B2 =

Correct option is (c). 11. For H-atom TH = 2pn3

B1 = 32 B2

Hence magnatic field increases 32 times.

8. For H-atom Lyman series is given by 1 1 1 = R é 2 - 2 ù for first line n = 2 êë1 l n úû 1 1 Þ = R é1 - 2 ù êë l 2 úû 1 3R Þ = l 4 h Momentum of photon Pp = l

2pn3 z2

For H-atom in ground state TH = 2p For H-like atom in first excited state (2p) ´ 23 2p ´ 8 = z2 z2 2 ´ 2p ´ 8 But TH = 2Tx Þ 2p = z2 Tx =

Þ

z2 = 16

Þ

z=4

Hence correct option is (c).

Let momentum of atom pA Q Initial momentum was zero. Hence using momentum conservation law, we get h 3 hR pA = pB Þ Mv = = l 4 3 hR Þ v= 4M Hence the correct option is (a). 9. Light wave equation is 200 V/m sin (1.5 ´ 1015 sec-1) t ´ cos (0.5 ´ 1015 sec-1) t

Maxmum incident energy

and for H-like atom Tx =

The correct option is (d).

Here maximum frequency =

10. Since in Balmer series of H-like atom wavelengths (in visible region) are found same or smaller hence the gas was initially in second excited state.

1.5 ´ 1015 2p

12. For K a line of X-ray 1 a2 = ( z - 1)2 l c Q z (atomic No.) for Pb204 , Pb206 , Pb208 are same hence l1 = l2 = l3. Hence correct option is (c). 13. The correct option is (d). -13.6 eV 14. Since En = n2 Þ E1 = - 13.6 eV and first excited state E2 =

-13.6 eV 4

77 E2 = - 3.4 eV

Þ

Þ z2 =

DE = E2 - E1 = 10.2 if K < 10.2 eV The electron collide elastically with H-atom in ground state. The correct option is (c). 1 1 1 15. For Lyman series = R é 2 - 2 ù here n = 3 l n ûú ëê1 1 1 = R é1 - ù l 9 ûú ëê

Þ

Pphoton = But

Þ

2 ´ 1836 ´ 9.1 ´ 10-31 ´ 1.6 ´ 10-19 ´ V V = 8.15 ´ 104 volt

n

-34

8 ´ 1.097 ´ 10 ´ 6.6 ´ 10 9 ´ 1837 ´ 9.1 ´ 10-31

Þ v = 4 m/s Hence correct option is (a). 16. Power = VI = 150 ´ 103 ´ 10 ´ 10-3 = 1500 W The 99% power heated the target hence 99 Heating power = ´ 1500 = 15 ´ 99 W 100 The rate at which target is heated per sec. (in cal) 15 ´ 99 ~ éQ1J = 1 cal ù = - 355 êë úû 4.2 4.2

The correct option is (d). eu r 1 20. Since m n = n n Q un µ and rn µ n2 2p n Þ m n = kn Where k is constant for H-atom For ground state m 1 = k ´ 1 = k

13.6 eV ´ z2 n2 13.6 eV 2 -13.6 eV 2 E3 = z and E4 = z 9 16 1 1 DE = E4 - E3 = (13.6) eV ´ z2 é - ù êë9 16 úû

17. En = -

13.6 (eV) ´ 22 ´ 7 = 32.4 eV 16 ´ 9

…(i)

For third excited state n = 4 m2 = k ´ 4 = 4 k

…(ii)

From Eqs. (i) and (ii) we get m 2 = 4m 1 Hence correct option is (d). 21. By conservation of momentum M H v = ( M H + M H ) v¢ Þ v¢ =

Hence correct option is (c).

Þ DE =

6.6 ´ 10-34

Hence correct option is (b). 1 19. Since En µ 2 and Ln µ n n 1 Hence En µ L 2

h 8 Rh = l 9

7

Þ v=

Hence correct option is (d). h h h 18. l = = = p 2m p K 2m p eV

Þ

PPhoton = PH- atom 8 Rh = Mp ´ v 9

Þ

Þ z =7

10-13 =

1 8R = l 9

Þ

16 ´ 32.4 ´ 9 = 49 13.6 ´ 7

v 2

Let initial KE of H-atom = K K Final KE of each-H-atom = 2 K -13.6 For excitation = E2 - E1 = + 13.6 2 4 K Þ = 10.2 eV 2 Þ

K = 20.4 eV

Hence correct option is (a).

78 22. We know that for H-like atom

Þ lmin =

En = - K n Þ K n = 3.4 eV h h l= = p 2me K =

Hence correct option is (b).

2 ´ 9.1 ´ 10-31 ´ 3.4 ´ 1.6 ´ 10-19

Þ l = 6.6 Å Hence both options (a) and (b) are correct.

1.1 e =

hc -W l2

27. Photo emission will stop when potential of sphere becomes stopping potential 1 q Þ =2V 4 pe 0 r Since

…(i) …(ii)

Subtracting Eq. (i) from Eq. (ii) we get é1 ù 1 0.5 e = hc ê -10 ú l 4950 ´ 10 ë 2 û Þ

0.5 ´ 1.6 ´ 10-19 1 1 + = 6.62 ´ 10-34 ´ 3 ´ 108 4950 ´ 10-10 l2

From Eq. (i) 6.6 ´ 10-34 ´ 3 ´ 108 ~ W = - 1.9 eV 4.95 ´ 10-7 ´ 1.6 ´ 10-19

(KE) max = eV0

hence Þ

V0 = 2 q = 8 pe 0 r coulomb

Hence correct option is (b) 28. Let t be the time for photo emission 1 q ´ ´ t =2 4 pe 0 r t=

8 pe 0 r q

Intensity of light at 0.8 from source I=

Þ l2 ~ - 4111 Å

3.2 ´ 10-3 ~ - 4 ´ 10-4 W/m 2 2 4 p ´ (0.8)

Energy incident on the sphere in unit time E1 = p ´ (8 ´ 10-3)2 ´ 4 ´ 10-4 = 8.04 ´ 10-8 W Energy of each photon E2 = 5 ´ 1.6 ´ 10-19 = 8 ´ 10-19 J

23. W = 1.9 eV Hence correct option is (c). 24. l = 4111 Å

Total number of photons incident on the sphere per second h=

Hence correct option is (c) 25. Since magnatic field does not change the KE of electrons hence retarding potential remain same. Hence correct option is (c). 26. (KE) max = 5 eV - 3 eV = 2 eV

lmin

2 ´ 9.1 ´ 10-31 ´ 2 ´ 1.6 ´ 10-19 = 8.69 Å

6.6 ´ 10-34

Hence answer is (c) both are correct. hc 23-25. 0.6e = -W 4950 ´ 10-10

6.6 ´ 10-34

= 2 ´ 1.6 ´ - 19 J h h = = p 2me K

E1 8.04 ´ 10-8 = = 1011 E2 8 ´ 10-19

Since 106 photons emit one electron. Hence the total number of photoelectron n 1011 per sec is n2 = 11 = 6 = 105 10 10 Therefore, q = n2 ´ e ´ t = 105 ´ 1.6 ´ 10-19 ´ t Þ

9 ´ 109 ´ 1.6 ´ 10-14 ´ t =2 r

79 t=

2 ´ 8 ´ 10-3 Þ t = 111 s 9 ´ 1.6 ´ 10-5

and there are three transitions shown as (1), (2) and (3) belonging to Lymen series.

Hence correct option is (c). ¢

Hence correct options are (a), (b) and (d).

More than one options are correct

6. Q f = a ( z - b)

hc if v increases l0 decreases eV hence the interval between lKa and l0 as well as lKb and l0 increases.

f versus z is a straight line

1. Since l0 =

Q hence

The correct options are (b) and (c). 1 1 2. R µ n2 , V µ and E µ 2 for Bohr model of n n H-atom V Þ VR µ n and µn E Hence, the correct options are (a) and (c). 3. For Bohr model of H-atom L µ n, r µ n2 and T µ n3 rL is independent of n T L 1 T µ 2 and µ n, L µ n3 T n r

Hence

Hence correct options are (a), (b) and (c). h h 4. Q l = and l = mv 2mK Hence heavy particle has smallest wavelength when speed and KE both particle are same. The correct options are (a) and (c). 5. Since there are six different wavelength n=4 n=3 1 n=2 2

3 n=1

Hence, final state will be n = 4. Since two wavelengths are longer than l0 [(From n = 4 ® 3 and n = 3 ® 2)] Hence initial state was n = 2

f =

c Þ l

1 a = ( z - b) l c

1 versus z is a straight line l f = a2 ( z - b)

Þ log f = log a2 + log ( z - b) which is a straight line Hence correct options are (a), (b) and (c). ¢

Match the Columns :

1. Lymen series lies in UV region, Balmer series lies in visible region and Paschen and Brackelt series lie in infrared region. Hence (a)

r

(b)

q

(c)

p

(d)

p

-13.6 eV n2 -13.6 -13.6 E2 = eV = eV 2 4 2

2. For H-atom En = Þ

Ionization energy from first excited state of H-atom 13.6 …(i) E = - E2 = eV 4 For He + ion - (13.6) eV ´ Z 2 for He + z = 2 n2 - 13.6 eV Þ En (He) = ´4 n2 EH (He) =

Ionization energy of He + atom from ground state = (13.6) eV ´ 4 = 4 E ´ 4 from Eq. (i) = 16 E

80 E2 (He) =

- (13.6) eV ´ 4 = - (13.6) eV 4

But E2 = - K 2 Þ K 2 = (13.6) and U = - 2 K = - 2 ´ (13.6) = - 2 ´ 4 E = - 8 E From Eq. (i) KE in ground state of He + ion = (13.6) eV ´ 4 = 4 E ´ 4 = 16E Ionisation energy from Ist executive state æ -13.6 eV ö = -ç ÷ ´ 4 = 13.6 eV = 4E 4 è ø Hence correct match are

3. K max

(a) ¾®

s

(b) ¾®

r

(c) ¾®

s

(d) ¾®

p

h W = hn - W and V0 = n e e

1 1 1 = R é 2 - 2 ù n = 3, 4, 5 … ê lB n úû ë2 For 2nd line n = 4 and lB = l 1 æ1 1 ö Þ =ç ÷R l è 4 16 ø 16 Þ l= 3R 1 1 1 =Ré 2 - 2ù ê lB 3 úû ë2 1 lB =

Þ

lB

1

1

Þ (a)® p R ´ (25 - 4) 21R 1 1 1 =Ré 2 - 2ù = = ê ú lB 25 ´ 4 100 5 û ë2 3

Þ lB = 3

Slope of line -1 is h, Y1 = W h W Slope of line -2 is , Y2 = e e Hence correct match are (a) ¾®

q

(b) ¾®

p

(c) ¾®

r

(d) ¾®

s

2pr z n2 4. T = Qr µ ÞUµ v z n Þ

Tµ

n3 z2

Q Q Q

For Lyman series 1 1 1 = R é 2 - 2 ù n = 2, 3, 4 êë1 lL n úû Þ

1 1 1 =Ré 2 - 2ù ê lL 2 úû ë1 1

1

L µ n Þ (b) ® s z V µ Þ (c) ® s n

5. Balmer series is

4 3R 5 l Þ (c) ® q = = æ 16 ö 4 ç ÷ è l ø 1 1 8R = R é1 - ù = lL 9 ûú 9 ëê

Þ lL = 1

n2 Þ (d) ® q Rµ z

100 100 = 21 R 21 ´ æ 16 ö ç ÷ è 3l ø 100 ´ 3 l 25l = = 21 ´ 16 28

Þ (b)® ( s)

Þ lL

Þ ( a) ® r

36 ´ 3 l 36 36 = = 16 5R 5 æ ö 16 ´ 5 ç ÷ è 3l ø 27 ö = æç ÷l è 20 ø

Þ

2

Þ lL = 2

9 ´ 3 l 27 l 9 9 = = = 8 R 8 ´ 16 16 ´ 8 128 3l

Þ (d) ® s

81 6. The proper match are (a)® s Q X-ray is inverse process of photoelectric effect [high energy electrons convert in electromagnetic radiation] Þ (b) ® p Q

lc µ

1 Þ (c)® q V

Q Wavelength of continuos X-ray depends on voltage Þ (d) ® q 7. Q (KE) max µ f and stopping potential µ f Þ (a) ® p, r

Q stopping potental µ f hence it remains same Þ (b)® s Q Current is directly proportional to Intensity. Hence sat current increases but stopping potential does not chase Þ (c) ® q, s Q(KE) max = hf - W and stopping potential æ hf W ö =ç ÷ e ø è e If W is decreased (KE) max and stopping potential increased Þ (d) ® p, r

Modern Physics II Introductory Exercise 30.1 1. R0 = lN 0 = 8000 Bq

Þ

l=

0.223 4 ´ 3600

R1 = lN = 1000 Bq R1 1 N = = = e- lt Þ log 8 = lt R0 8 N 0

Þ

Þ

log 2 3 log 2 = ´9 T1/2

(b) R0 = 10 mci = 10 ´ 3.7 ´ 1010 ´ 10-3 Bq

Þ

9 = = 3 days 3

Þ

T1/2

= 3.7 ´ 108 Bq R0 = lN 0

Average, life = 1.44 ´ T1/2 = 1.44 ´ 3 = 4.33 days 2. R0 = lN 0 0.693 ´ N0 64.8 4

N0 =

40 ´ 64.8 ´ 3.7 ´ 10 0.693

= 13.83 ´ 107 Now,

3.7 ´ 108 R Þ N0 = 0 = l 1.55 ´ 10-5 = 2.39 ´ 1013 (atoms)

Þ 40 ´ 3.7 ´ 1010 ´ 10-6 = Þ

T1/2

l = 1.55 ´ 10-5 /s 0.693 0.693 = = = 12.4 h l 1.55 ´ 10-5

N = N0 e

(c) R = R0 e- lt = 10 mCi ´ e

- 0.693 ´ 10

N10 = N 0 e

64.8 - 0.693 ´ 10

= 13.83 ´ 107 ´ e

64.8 - 0.693 ´ 12

N12 = 13.83 ´ 107 ´ e Þ

64.8

9

» 9.47 ´ 10 nuclei

Þ

R0 = 10 mCi, R = 8 mCi Þ log

10 = l ´ 4 ´ 3600 8

= 1.87 mCi

T1/2

R0 6 ´ 1011Bq = N0 1015 = 6 ´ 10-4 /s 0.693 0.693 = = = 1.16 ´ 103s l 6 ´ 10-4

5. N x = N y = N 0 T1/2 x = 50 min and T1/ 2 y = 100 min

N10 - N12 = 13.83 ´ 107 - 0.693 ´ 12 é - 0.693 ´ 10 ù - e 64.8 ú êe 64.8 êë úû

3. (a)

0.693 ´ 30 12.4

4. R0 = lN 0 Þ l=

- lt

-

R0 = e lt R

n

4

1 1 N N x = ( N 0 ) æç ö÷ = N 0 æç ö÷ = 0 16 è2ø è2ø n

2

1 1 N N y = N 0 æç ö÷ = N 0 æç ö÷ = 0 2 2 4 è ø è ø N0 Nx 1 = 16 = N y N0 4 4

83

Introductory Exercise 30.2 1.

2

- 4.002602) u

DE = DMc

Þ DM =

DM = 0.0004589 u

P = 109 J/s = 109 ´ 24 ´ 60 J/day

Here,

109 ´ 24 ´ 60 = 9.6 ´ 10-4 kg (3 ´ 108 )2

DE = DM ´ 931.5 MeV = 0.0004589 ´ 931.5

9

2. Number of fission =

10 J/s 200 ´ 106 ´ 1.6 ´ 10-19

= 4.27 MeV 4. Complete reactions are (a) 3Li 6 + 1H2 ® 4Be 7 + 0 n1

19

= 3.125 ´ 10 3. Given reaction is

(b) 17 Cl 35 + 1H1 ® 16S 32 + 2He4

238 92 U

(c) 4 Be 9 + 2He4 ® 3 (2 He4 ) + 0 n1

¾® 90 Th234 + 2He4

DM = (238.050784 - 234.043593

(d)

35 Br

79

+ 1H2 ® 36Kr79 + 2 ( 0 n1)

AIEEE Corner ¢

Subjective Questions (Level 1)

Radioactivity 1. (a) Initially the rate of distingration is æ dN ö -ç ÷ = lN 0 è dt ø0 - dN After 5 min = lN dt æ dN ö ç ÷ N 0 è dt ø0 4750 Þ = = = 1.76 dN N 2700 dt æ N ö ÷÷ log çç è N0 ø Now N = N 0 e- lt or l = t 2.3026 N0 Þ l= log10 t N 2.3026 = log10 (1.76) 5 Þ l = 0.113 /min 0.693 0.693 (b) Half-life = = = 6.132 min l 0.113

2. We have A = lN 6 ´ 1011 = l ´ 1 ´ 1015 l = 6 ´ 10-4 s -1 0.693 0.693 T1/2 = = = 1155 s l 6 ´ 10-4

Þ

= 19.25 min 3. A = lN A = 8 Ci = 8 ´ 3.7 ´ 1010 decay/s 0.693 0.693 l= = T1/2 5.3 yr =

0.693 5.3 ´ 365 ´ 24 ´ 60 ´ 60 s

Þ N= =

A l

8 ´ 3.7 ´ 1010 ´ 5.3 ´ 365 ´ 24 ´ 3600 0.693

Þ N = 7.2 ´ 1019

84 6.023 ´ 1023 nuclei = 60 g 60 g 1= 6.023 ´ 1023 Hence 7.2 ´ 1019 nuclei =

60 ´ 7.2 ´ 1019 = 7.11 ´ 10-3 g 6.023 ´ 1023

4. Number of decay per second m 1 0.693 ´ NA ´ l = ´ 6 ´ 1023 ´ M 238 4.5 ´ 109 yr =

6 ´ 1023 0.693 ´ 238 4.5 ´ 109 ´ 365 ´ 24 ´ 60 ´ 60 s

= 1.23 ´ 104 decay/s 5. Probability of decay P = (1 - e

- lt

Þ P = (1 - e

) = (1 - e - 5/10

- t/ T mean

) =1 - e

-2

)

= 0.39

( R2 ) tP

R = ( R1) tP

It means Hence

Þ Þ Þ 7. ( R1) 0 P

32

( R2 ) 0 P

33

238 1 of original U nuclei decays. 4 N = N 0 e - lt N ö æ -l ç N0 - 0 ÷ = N0 e 4 ø è

( R1) tP

32

+ ( R2 ) tP 33 log 2 ´ 60 ´ 365 e 14

32

+ l2 e

-

log 2 ´ 60 ´ 365 ù 25 ú

úû

log 2 ´ 60 ´ 365 log 2 é ´ 60 ´ 365 ù 14 ê4 l1 e ú + l2 e 25 êë úû 3 mCi = é 4 ´ log 2 log 2 ù + ê 14 25 úû ë log 2 é 4 log 2 - log 2 ´ 60 ´ 365 log 2 ´ 60 ´ 365 ù 14 ê ú e + ´ e 25 25 êë 14 úû

» 0.205 mCi 8. Complete reactions are 226 ¾¾® a + 86RN222 88 Ra (b) 8 O19 ¾¾® 9F19 + e + n (c) 13 Al25 ¾¾® 12Mg25 + e+ + n

(a)

9. Only reaction (b) is possible. 10. DE = (7 ´ 1.000783 + 7 ´ 1.00867 - 14.00307)

11. DE = [8 m p + 8 mn - m ( 8 O16 )] ´ 931.5

t = 1.88 ´ 109 yr

12. (a) Number of nuclei in kg

´ 931.5 MeV Þ DE = 104.72 MeV

= (8 ´ 1.007825 + 8 ´ 1008665 . - 15.994915) ´ 931.5 = 127.6 MeV

= l1 4 N 0

=

= l2 N 0 32

+ ( R2 ) 0 P

= l1 N = l1 4 N 0 e- l1 t = l1 ´ 4 N 0 e

-

6.023 ´ 1023 ´1 235

Energy 33

= N 0 ( 4 l1 + l2 ) = 3 mCi 32

log 2 ´ 60 ´ 365 25

3 = e - lt 4 log 4 - log 3 t= l (log 4 - log 3) t= ´ 4.5 ´ 109 yr 0.693

R0 = ( R1) 0 P ( R1) tP

-

é = N 0 ê4l1 êë 3 mCi = ( 4 l1 + l2 )

206

6. Since initially no Pb nuclei is present and Nv after time t the ratio of =3 N Pb

33

= l2 ´ N 0 e

log 2 ´ 60 ´ 365 14

=

6.023 ´ 1023 ´ 200 ´ 106 ´ 1.6 ´ 10-19 235

= 8.09 ´ 1013 J

85 (b) Mass =

8.09 ´ 1013 J 30 ´ 103 J/g

= [2 (2.014102) - 3.016049 - 1.007825] u DM = 0.000433 u

13

=

8.09 ´ 10 g 30 ´ 103

Q = DM ´ 931.5 = 0.000433 ´ 931.5 » 4.05 MeV

8.09 1 = ´ 109 ´ 3 kg 3 10

(b) DM = (2 ´ 2.014102 - 3.016049 - 1.008665) ´ 931.5

= 2.7 ´ 106 kg 13. Applying conservation of momentum

Þ

M a va = M TivTi M v vTi = a a M Ti Ka = K Ti =

DM = 17.57 MeV

1 1 M v2 = 2 Ti Ti 2

K Ti

Þ

M 2 vL M Ti a2 a M Ti

æ 1 M v2 ö ç a a÷ è2 ø

4 ´ 6.802 MeV 208 1 = ´ 6.802 = 0.1308 MeV 52 14. Power = 100 MW = 108 W = 108 J/s 8

Energy per fission = 185 MeV 1021 / s. 1.6 ´ 185

Number of nuclei in 1 kg 6.023 ´ 1026 U235 = 235 Hence t =

6.023 ´ 1026 ´ 1.6 ´ 185 235 ´ 1021

= 8.78 days 15. (a) The given reaction is 2 1H

He4 + He4 ¾® Be 8 DM = (2 ´ 4.0026 - 8.0053) u = (8.0052 - 8.0053) u Q DM is negative this reaction is not energetically favourable DE = DM ´ 931.5 = - 1 ´ 10-4 ´ 931.5 MeV = - 93.15 keV

21

10 10 MeV = 1.6 s 1.6 ´ 10-13 J/MeV

Hence number of fissions =

16. The given reaction is

DM = - (0.0001) u

=

=

(c) DM = (2.014102 + 3.016049 - 4.002603 - 1.008665) ´ 931.5

1 M a v2a = 6.802 MeV 2

M = a M Ti

DM = 3.25 MeV

17. Number of nuclei in 1 kg water =

6.023 ´ 1026 1.5 ´ 10-2 ´ 18 100 6.023 ´ 1.5 = ´ 1022 18

Heavy water =

Energy realesed per fission = (2 ´ 2.014102 - 3.016049 - 1.007825) ´ 931.5 = 4.33 ´ 10-3 ´ 931.5 ´ 106 ´ 1.6 ´ 10-19 J Hence total energy =

+ 1H2 ¾® 1H 3 + 1H1

DM = [2m (1H2 ) - m (1H 3) - m (1H1)]

6.023 ´ 1026 18

6.023 ´ 1.5 ´ 1022 18

´ 4.33 ´ 931.5 ´ 1.6 ´ 10-16 = 3200 MJ

86 ¢

Objective Questions (Level-1)

1. Since during b- decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino as n ® P + e- + g.

n

1 1 1 But N = N 0 æç ö÷ Þ N 0 = N 0 æç ö÷ 8 è2ø è2ø Þ Þ

Hence Correct option is (c). 2. Since nuclear force is same for all nucleons. Hence F1 = F2 = F3 Correct option is (a).. 3. Given reaction is

90

X

200

¾® 80 Y

168

Difference in mass number = 200 - 168 = 32 32 Hence Number of a-particles = =8 4 Difference in atomic number = 10 hence number of b -particles = 6 4. The reaction is + 0 n1 ¾® 54 Xe138 +

38 Sr

94

+ 3 ( 0 n1)

The correct option is (b) three neutrons. 5. The reactions are A ® b + a and b ® C + 2b After one a atomic number reduced by 2 and after 2b atomic number increased by. Hence A and C are isotopes Correct option is (d). 6. Here m p = 1.00785 u, mn = 1.00866 u and ma = 4.00274 u Dm = 2( m p + mn ) - ma Þ Dm = [2 (1.00785 + 1.00866) - 4.00274 ] u Þ Dm = 00 . 3028 u DE = Dm ´ 931.5 MeV = 0.3028 ´ 931.5 = 28.21 MeV Hence correct option is (c). 7 1 7. N = N 0 - N 0 = N 0 8 8

n =3 3 ´ T1/2 = 8 s 8 T1/2 = s 3

Hence correct option is (d). 8. N = N 0 e- lt for mean life t = Þ N = N0 e

-l ´

1 l

=

1 l

N0 e

Hence the fraction disintegrated N -N æ 1 = 0 = ç 1 - ö÷ N0 eø è Correct option is (b). 7 1 9. N = N 0 - N 0 = N 0 8 8

Hence correct option is (d). 235 92 U

Þ

n

1 But N = N 0 æç ö÷ è2ø

n

n

Þ

1 æ1ö = ç ÷ Þn = 3 8 è2ø

Hence half life T1/2 =

15 min = 5 min 3

Correct option is (a). 10. Since radioactive substance loses half of its activity in 4 days it means its half life T1/2 = 4 days Now A = 5% of A0 1 Þ A= A0 20 A 1 Þ = A0 20 But Þ

1 = e - lt 20 log 20 log 20 = lt Þ t = l A = A0 e - l t Þ

87 But l =

log e 2 T1/ 2

Þ t = T1/2

log e 20 log10 20 = T1/ 2 = 4.32 ´ 4 log e 2 log10 2

Þ t = 17.3 days Hence correct option is (c). 11. Total energy released per sec = 1.6 MW = 1.6 ´ 106 J/s Energy released per fission= 200 MeV 6

-19

= 200 ´ 10 ´ 1.6 ´ 10 -11

= 2 ´ 1.6 ´ 10

J

1.6 ´ 106 = 5 ´ 1016 /s 2 ´ 1.6 ´ 10-11

Hence correct option is (a). 12. Q R = R0 A Volume

Þ

– N =e N0

1/ 3

4 4 = pR 3 = pR03 A 3 3

Mass of nucleus = A ´ 1.67 ´ 10-27 kg A ´ 1.67 ´ 10-27 kg mass Density r = = 4 volume pR03 A 3 1.67 ´ 10-27 kg Þ r= 4 pR03 3 Qr is independent of A, hence ratio of r densities 1 = 1. r2

N = e – lt N0

0. 693 ´10 6. 93

= e –1 =

1 e

Fractional change N -N æ 1 = 0 = ç 1 - ö÷ ~ - 063 . N0 eø è Hence correct option is (b). 14. Since radioactive substance reduce to about N 6% it means N = 0 16

J

Number of fission per second =

N = N 0 e – lt Þ

13.

n

We have

1 N = ( N 0 ) æç ö÷ è2ø

Þ

1 æ1ö =ç ÷ Þ n = 4 16 è 2 ø

n

4 ´ T1/2 = 2 h Þ T1/2 = 30 min Hence correct option is (a). 15. Probability of a nucleus for survival of time t. p(survival) =

N e - lt N = 0 = e - lt N0 N0

For one mean life t = Þ Psurvival = e

-l ´

1 l

1 l

= e -1 =

1 e

Hence Correct option is (a).

Correct option is (d). ¢

Assertion and Reason

1. Here both assertion and reason are true but reason does not explain assertion. Hence correct option is (b).

3. Here both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b).

2. Here assertion is false but reason is true since for heavier nucleus binding energy per nucleon is least.

4. Here a ssertion is true but reason is false since electromagnetic waves are produced by accelerating charge particles.

Correct option is (d).

Correct option is (c).

88 5. Here assertion is wrong since b-decay –

9. Here reason is true but assertion is false

process is n ® p + e + v

Q 1 amu = 931.5 MeV

but reason is true hence correct option is (d).

Correct option is (d)

6. Here assertion is true but reason is false. Correct option is (c). 7. Here both assertion and reason are true and reason may or may not be true. Correct option is (a, b) 8. Both assertion and reason are true but reason is not correct explanation of assertion. Hence correct option is (b).

10. Both assertion and reason are true but reason does not correctly explain assertion. Hence correct option is (b). 11. Here both assertion and reason are true and reason may or may not be correct explanation of assertion . Hence correct options are (a, b).

Objective Questions (Level 2) ¢

Single option correct

1. Let initially substance have N i nuclei then N = N i e - lt dN = - l N i e - lt dt At

t=t

we get æ dN ö = - l N i e - lt = N 0 ç ÷ è dt øt = t At

…(i)

From Eqs. (ii) and (iii) N = 0 64 Hence, correct option is (b). log 2 log 2 2. We have l = l1 + l2 = + 30 60 log 2 Þ l= 20 Now

- t log 2

t = 4t Þ

N æ dN ö = - l N i e -4 lt = 0 ç ÷ 16 è dt øt = 4 t

…(ii) Þ

Dividing Eq. (i) and Eq. (ii) we get e3lt = 16 11 Now at t = æç ö÷ t è 2 ø

…(iii)

Þ Þ

æ dN ö = - lN i e ç ÷ è dt øt = 11t

- 11l 2

- 8 lt

= - lN i e

2

- l N i e -4 lt e

3lt

- 3lt

´e =

N0 = N 0 e 20 4 t log 4 = log 2 20 t 2 log 2 = log 2 20 t = 40 yr

Hence correct option is (c).

t

2

=

N = N 0 e - lt

2

N0 16 ´ 16

3. From graph it is clear that number of nucleons in X is N 3 and binding energy per nucleon is E3 for Y nucleon is N2 and BE per nucleon is E2 . Hence X + Y = E3 N 3 + E2 N2 Similerly W = E1 N1

89 The reaction is W ® X + Y The energy released is ( E3 N 3 + E2 N2 - E1 N1)

8. By conservation of momentum M H v = ( M H + M H ) n¢ v v¢ = 2

Þ

Hence Correct option is (b). 4. Energy = (110 ´ 8.2 + 90 ´ 8.2 - 200 ´ 7.4) = 200 ´ (8.2 - 7.4) = 200 ´ 0.8 = 160 MeV Hence correct option is (d). 5. The reaction is 2 1H

+ 1H2 ® 2He4 + energy

Energy

= (4 ´ 7 - 2 ´ 1.1) MeV = (28 - 4.4) = 23.6 MeV

Hence Correct option is (b). 6. Total energy released per second = 16 ´ 106 W = 16 ´ 106 J/s Energy per fission = 200 MeV = 200 ´ 106 ´ 1.6 ´ 10-19 = 2 ´ 1.6 ´ 10-11 J Q Efficiency = 50% Hence power (energy converted per second) 50 = 2 ´ 1.6 ´ 10-11 ´ = 1.6 ´ 10-11 J 100 Number of fission =

16 ´ 106 = 1018 /s 1.6 ´ 10-11

Hence correct option is (d). dN 7. = A - lN dt Q After time N become dN conservation Þ =0 dt A A AT Þ N= = = l log 2 log 2 T Hence correct option is (d).

Let initial KE of H-atom = K K Final KE of each H-atom = 2 For excitation K æ - 13.6 ö = E2 - E1 = ç + 13.6 ÷ eV 2 è 4 ø K Þ = 10.2 eV 2 Þ

K = 2 ´ 10.2 ´ 1.6 ´ 10-19

Þ

1 M u2 = 2 ´ 10.2 ´ 1.2 ´ 1.6 ´ 10-19 2 H

Þ uH =

2 ´ 1.2 ´ 10.2 ´ 2 ´ 1.6 ´ 10-19 1.673 ´ 1027

= 6.25 ´ 104 m/s Hence correct option is (c). 9. Let us suppose just before the death no radioactive atoms were present hence original activity A0 is given as A0 = l N 0

…(i)

After death the radioactivity decreases exponentially ie, dN …(ii) A= = l N = l N 0 e - lt dt Dividing eq. (ii) by eq. (i) we get A = e - lt A0 or lt = log

A0 1 A or t = log 0 A l A

Now A0 = 15 decay/min/gram 375 decay/min/g A= 200 0.693 but l= 5730 yr

90 15 ´ 200 5730 5730 200 ´ log = ´ log 0.693 375 0.693 25 5730 Þ t= ´ log 8 0.693 5730 = ´ 3 log 2 = 5730 ´ 3 0.693 t=

t=

1.29 ´ 1028 s » 1012 s. 1016

Hence correct option is (c). - log 2

12. N1 = N 0 e

- l1 t

…(i)

N Q = N 0 e - lt

and Now

AP = lN P and A Q = lN Q AP N = P = e- lt1 AQ N Q

Þ

æ AQ ö ÷ lt1 = log ç ç A ÷ è P ø

Þ

æ AQ ö æA ö 1 ÷ = T log ç Q ÷ t1 = log ç ç ÷ ç A ÷ l è AP ø è P ø

Þ

and

…(i)

11. The given reactions are

Þ 31H2 ® 2He4 + n + p

t

…(ii)

R1 = l1 N1

…(iii)

R2 = l2 = l2 N2

…(iv)

Þ

l1 N1 ´ =1 l2 N2 é t 2 - t1 ù ú´t t1 t 2 û

log 2 ê l1 N2 ë = =e l2 N1

Þ Þ Þ

log

æt -t ö t2 = t ´ log 2 ´ çç 2 1 ÷÷ t1 è t1t2 ø t1 =

+ 1H2 ® 1H 3 + p + 1H 3 ® 2He4 + n

t2

Let after time t, R1 = R2 then R1 Þ =1 R1

Hence correct option is (b). 2 1H 2 1H

t

N2 = N 0 e- l2 t = N 0 e

Hence correct option is (c). 10. N P = N 0 e- l(t1 + t)

t1

- log 2

t = 17190 yr

Þ

= N0 e

t1t2 t ´ log 2 0.693 ( t2 - t1) t1

Hence Correct option is (a). 13. The given reaction is

Z

X 232 ¾®

X 232 ¾® 90 Y A + a 90 Y

A

+ 2He4

Þ

Z

Dm = (3 ´ 2.014 - 4.001 - 1.007 - 1.008)

Þ

Z = 92 and A = 228

Dm = 0.026 amu

Q Initially X is in rest hence momentum of a-particle after decay will be equal and opposite of Y .

Mass defect

Energy Released = 0.026 ´ 931 MeV = 3.87 ´ 10-12 J This energy produced by the three deutronatoms. Total energy released by 40

10

=

deutrons 1040 ´ 3.87 ´ 10-12 J = 1.29 ´ 1028 J 3

The average power P = 1016 W = 1016 J/s Therefore total time to exhaust all deutrons of the star will be

Þ Þ

M Y vY = M a va M vY = a va MY

Total kinetic energy 1 K T = ( M a v2a + M Y v2Y ) 2 Þ

KT =

1é M 2a v2a ù 2 êM a va + M y ú 2ë M 2Y û

91 é Ma ù ê1 + ú M Y û ë 4 ù K T = K a é1 + êë 228 úû 232 Ka = K 228 T KT =

Þ Þ Þ

1 M a v2a 2

Hence Correct option is (b). 14. Energy of emitted photon = 7 MeV 6

-19

= 7 ´ 10 ´ 1.6 ´ 10 -13

= 11.2 ´ 10

Momentum of photon = =

J 11.2 ´ 10-13 J 3 ´ 108 m/s

®

O = P nuc + P photon Þ P nuc = - P photon ®

®

K nuc = Þ K nucc =

2m

-42

=

11.2 ´ 11.2 ´ 10 Joule 9 ´ 2 ´ 24 ´ 1.66 ´ 10-27

11.2 ´ 112 . ´ 10-42 eV 18 ´ 24 ´ 1.66 ´ 10-27 ´ 1.6 ´ 10-19

» 1.1 keV Hence correct option is (b). 15. Let time interval between two instants is t1 then N1 = N 0 e- l (t + t1 )

Þ

t1 =

æ A ö 1 log çç 2 ÷÷ l è 2 A1 ø

Þ

t1 =

æ A ö T log çç 2 ÷÷ log 2 è 2 A1 ø

2 1H

+ 1H2 ¾® 1H 3 + 1H1

DM = [2m (1H2 ) - m (1H 3) - m (1H1)] = (2 ´ 2.014102 - 3.016049 - 1.007825) amu

Hence correct option is (c).

2 Pnuc = 2mK nuc 2 Pnuc

æ A ö log çç 2 ÷÷ = lt1 è 2 A1 ø

= 4 MeV

= 24 ´ 1.66 ´ 10-27 kg

Þ

Þ

DE = 4.33 ´ 10-3 ´ 931.5 MeV

11.2 ´ 10-21 kg-m/s 3

Mass of nucleus = 24 amu But

2 A1 = e- lt1 A2

= 4.33 ´ 10-3amu

Þ |P nuc|=|- P photon| Þ Pnuc =

Þ

16. The given reaction is

Applying conservation of momentum principle ®

Þ

A2 = N2 l = l (2 N 0 ) e- lt A1 1 = ´ e- lt1 A2 2

Hence Correct option is (c).

Q Initial nucleus is stationary

®

N2 = 2 N 0 e- lt A1 = N1l = lN 0 e- l (t + t1 )

J

11.2 ´ 10-21 kg-m/s 3

®

and

17. Number of fusion required to generate 1 kWh =

1 ´ 103 ´ 3600 4 ´ 106 ´ 1.6 ´ 10-19

=

36 ´ 1018 = 5.6 ´ 1018 » 1018 6.4

Hence correct option is (b). 18. The energy released = 4 MeV This energy produced by two atoms. Hence energy produced per atom = 2 MeV = 2 ´ 1.6 ´ 10-13 J Hence number of atom fused to produced 1 kWJ

92 =

36 ´ 105 18 = ´ 1018 2 ´ 1.6 ´ 10-13 1.6

Mass of deutrium which contain 18 ´ 1018 atom 1.6 ¢

=

18 ´ 1018 = 3.7 ´ 10-5 kg. 1.6 ´ 6.02 ´ 1023

Hence correct option is (c).

More Than one Option is Correct 4

1. x = N 0 , y = lN 0 Þ

x N0 1 = = y lN 0 l

where l is decay constant x Hence is constant throught. y Q

Þ

x 1 1 T = = = y l 0.693 0.693 T x > T, xy = l( N 0 )2 y

For one half life N =

N0 2

2

Þ ( xy) T

1 N N x = N 0 æç ö÷ = 0 2 16 è ø

Þ

lN 20 xy æN ö = lç 0 ÷ = = 4 4 è 2 ø

Hence correct options are (a), (b) and (d). 2. The correct options are (a), (b), (c) and (d). 3. A nucleus in excited state emits a high energy photon called as g-ray. The reaction is X * ¾® X + g Hence by gamma radiation atomic number and mass number are not changed. Since after emission of one a atomic number reduced by 2 (2 a4 ) and after 2b atomic number is increased by (2). Hence correct options are (a), (b) and (c). 4. Here half lives are T and 2T and N x = N 0 , N y = N 0 after 4T for first substance = 4 half lives and after 4T the second substance = 2 Half lines.

2

1 N N y = N 0 æç ö÷ = 0 4 è2ø N /16 1 N x= x = 0 = Ny N 0 /4 4

Þ

Let their activity are Rx and R y . Rx = lx N x and R y = l y N y 0.693 Rx lx N x 1 y= = ´ = T ´ R y l y N y 0.693 4 2T Rx 1 =y= Ry 2

Þ Þ

Þ

Hence correct options are (b) and (c). 5. Since nuclear forces are vary short range charge independent, no electromagnetic and they exchange (n ® p or p ® n). Hence the correct options are (a), (b), (c) and (d). 6. Q R = R0 A1/ 3 r=

A ´ 1.67 ´ 10-27 kg M = 4 4/3 pR 3 pR03 A 3

Þ r is independent of A. But r =

1.67 ´ 10-27 kg 4 ´ 3.14 ´ (1.3 ´ 10-15 ) 3 3

= 1.8 ´ 1017 kg/m 3 Hence correct options are (b) and (c).

93 ¢

Match the Columns

1. Here N 0 = x, let l is the decay constant. Þ N = N 0 e - lt = x e - lt dN dN = lx e- lt but dt dt

=y t=0

Þ

Q Binding energy increases when two or more lighter nucleus combine to form haviour nucleus. Hence correct match is

y = lx e 0 = lx y l= x

Þ

2. In reaction P + P ® Q energy is released

(a) ¾® p Similerly for reaction

Hence

P + P + R®R

(a) ¾® s

Correct match is

Half life T1/2 =

log 2 log 2 x = = log 2 y l y x

for reaction P + R = 2Q from graph BE per nucleon increases.

Hence (b) ¾® p We have activity R = lN = l xe- lt 1 at t = l R = lx

1 -l ´ l e

= l x e -1 =

lx e

but lx = y y R= e

Þ

Hence energy is released. Correct match is (c) ¾® p For the reaction P+Q=R We will check energy process if BE per nucleon is given. Hence data is not sufficient correct match is (d) ¾® s

Hence (c) ¾® r 1 Number of nuclei after time t = l N =xe

(b) ¾® p

-l ´

1 l

=

x e

Hence (d) ¾® s Thus correct match is

3. Since A and B are radioactive nuclei of ( A + B) decreases with time. Hence correct match is (a) ¾® q Q A is converted into B and B is converted into C and decay rate of A ® B and B ® C are not known. Hence correct match is (b) ¾® s

(b) ¾® r

Since at time passes A is converted to B and B is converted to C. Hence nuclei of ( B + C) increases. Correct match is

(c) ¾® r

(c) ¾® p

(d) ¾® s

Similerly the correct match for (d) is

(a) ¾® s

94 (d) ¾® s 4. After emission of 1 a particle mass no decreased by 4 but after emission of 1b particle atomic number will increase or decrease by 1. Hence for (a)

(a) ¾® p Since BE per nucleon of heavy nuclei is about 7.2 MeV. Hence (b) ¾® s

(b) ¾® p, r

X-ray photon have wavelength about 1 Å the energy of this wavelength is of order of 10 keV.

(c) ¾® s

(c) ¾® r

(d) ¾® q, r

Q Visible light energy of order of » 2 eV. Hence

(a) ¾® p, s

5. For

(d) ¾® q