# Physics Disha

October 23, 2017 | Author: akalikar | Category: Photoelectric Effect, Collision, Prism, Photon, Electron

Physics Disha...

#### Description

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PHYSICS I. HEAD-ON ELASTIC COLLISION If colliding bodies before and after collision remain in a line, the collision is said to be head-on collision. This will happen when bodies move along the line joining their geometric centres. r r r r Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 (u1 > u 2 ) r r along the same straight line. Let after collision their velocities become v1 and v2 in the same initial direction. Then

according to conservation of linear momentum, we have r r r r m1u1 + m2u 2 = m1v1 + m2 v 2 . Since all the colliding bodies before and after collision remain in the same line, so we can drop the vector signs from them. Thus we can write m1u1 + m2u2 = m1v1 + m2v2 … (i) or m1(u1 – v1) = m2(v2 – u2) …(ii) As kinetic energy before collision = kinetic energy after collision 1 1 m1u12 + m2u22 2 2 or m1(u12 – v12) or m1(u1 + v1)(u1 – v1) Dividing equation (iv) by (ii), we get u1 + v 1

\

or

u1 - u2

1 1 m1v12 + m2 v22 2 2 = m2(v22 – u22) = m2(v2 + u2) (v2 – u2)

=

…(iii) (iv)

= v 2 + u2 = v2 - v1.

…(v)

Thus velocity of m1 w.r.t. m2 before collision = velocity of m2 w.r.t. m1 after collision. or Also we have,

and

velocity of approach

=

velocity of separation

v1

=

æ m1 - m2 ö æ 2m2 ö çè m + m ÷ø u1 + çè m + m ÷ø u2 …(vi) 1 2 1 2

v2

æ m2 - m1 ö æ 2m1 ö = ç m + m ÷ u2 + ç m + m ÷ u1 ... (vii) è 1 è 1 2ø 2ø

Special cases : (i)

When colliding bodies are of equal masses, let m1 = m2 = m. From equation (vi) and (vii), we get v1 = u2 and v2 = u1 Hence when two bodies of equal masses collide elastically, their velocities get exchanged.

2 (ii)

If m1 = m2 = m and u2 = 0, then

v1 = 0 and v2 = u1. (iii) When a light body collides with a massive stationary body. Here m1 > m2 and u2 = 0 \ v1 = u1 and v2 = 2u. Transfer of kinetic energy during collision: Kinetic energy transferred from projectile to the target DK = decrease in K.E. of projectile =

1 1 m1u12 - m1v12 . 2 2

DK K

=

1 m u 2 - 1 m v2 2 1 1 2 11 1 m u2 2 1 1

DK K

æv ö = 1- ç 1 ÷ . èu ø

Fractional decrease in K.E.

or

2

…(viii)

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Perfectly inelastic collision in 1-D Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along a straight line. They make perfectly inelastic collision. Let after collision, their common velocity becomes v, then by conservation of momentum, we have m1u1 + m2u2 = (m1 + m2)v \

v

=

é m1u1 + m2 u2 ù ê ú. ë m1 + m2 û

The loss of K.E. in collision

DK

æ1 2 1 2ö 1 2 = çè m1u1 + m2u2 ÷ø - (m1 + m2 )v 2 2 2 æ1 2 1 2ö 1 = çè m1u1 + m2u2 ÷ø - (m1 + m2 ) 2 2 2 æ m1u1 + m2 u2 ö çè m + m ÷ø 1

=

1 æ m1m2 ö (u1 - u2 ) 2 . 2 çè m1 + m2 ÷ø

The loss of K.E.. will appear as heat and sound.

2

2

3

General analysis of 1-D collision Newton's experimental law : Coefficient of restitution It is defined as; e = =

or

e =

velocity of separation velocity of approach v2 - v1 u1 - u2

év -v ù év -v ù - ê 2 1 ú = - ê 1 2 ú. ë u2 - u1 û ë u1 - u2 û

The value of e depends on materials of colliding bodies. The value of e can be e £ 1. (i) For perfectly elastic collision, e = 1. (ii) For perfectly inelastic collision, e = 0.

Note: The coefficient of restitution is a 1–D concept. Thus in problem involving oblique collision, 'e' is defined only along the line of collision. In the absence of tangential forces the collision in the perpendicular direction is taken as elastic. Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 along a line. Let the coefficient of restitution between the bodies is e. After collision their velocities become v1 and v2 respectively. Then we have, m1u1 + m2u2 = m1v1 + m2v2 … (i) and

e =

-

v1 - v2 . u1 - u2

… (ii)

Solving equations (i) and (ii), we get v1

=

æ m1 - em2 ö æ (1 + e)m2 ö … (iii) çè m + m ÷ø u1 + çè m + m ÷ø u2 1

and

2

1

2

æ m2 - em1 ö æ (1 + e)m1 ö çè m + m ÷ø u2 + çè m + m ÷ø u1 …(iv)

v2

=

mu

= mv1 + mv2

1

2

1

2

Special case If m1 = m2 = m and u1 = u, u2 = 0, then and

e =

-

v1 - v2 u-0

After solving above equations, we get

\

v1

=

u (1 - e) 2

v2

=

u (1 + e) 2

v1 v2

=

1- e . 1+ e

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II. ELECTRIC

DIPOLE

A system of two equal and opposite charges fixed at a small distance constitutes a dipole. If l is the distance between the charges + q and – q, then dipole moment is defined as : r r P = ql . It is a vector quantity and its direction is from negative to positive charge.

r r Potential due to an electric dipole : Consider a dipole AB of dipole moment, P = q l . We want to calculate the electric potential at a point P, at a distance r from the centre of the dipole. Let line r makes an angle q with the line of dipole. The electric potential due to the charges of the dipole at P Vp

From the figure, PA = r +

\

=

l l cos q and PB = r - cos q, assuming that l 2C \ So, a ray of light will not emerge out from the prism, if A > 2C.

...(7)

Totally reflecting prism The critical angle for glass-air interface is 42°. Thus if we make a prism in such a way, that light ray incident into it at an angle greater than critical, then it becomes totally reflecting prism. Such a prism may be right angled isosceles (45° – 90° – 45°). They can be used to deviate rays through 90° or 180°.

Erecting prism This is also the right-angled isosceles prism. In this case rays of light should be parallel to the hypotenuse. By doing so the rays invert themselves and an inverted object appears as erect.

V.

PHOTO-ELECTRIC EFFECT

When light of certain frequency is incident on a metal surface, electrons are ejected from themetal. Thisphenomenon iscalled photoelectric effect (PEE). Electrons ejected from the metal are called photoelectrons. The photoelectric effect was first observed by Heinrich Hertz in 1887.

Work function We know that metals have large number of free electrons. These electrons move freely inside the metal but can not come out from it due to attraction of the positive ions. Some energy is needed to liberate the electrons from the bondage of the attraction of the ions. The minimum energy required to liberate the electrons from the metal surface, is called work function, and is represented by W0.

Work functions of some photometals

11 Metal

Work function (ev)

Metal

Work function (eV)

Cesium

1.9

Calcium

3.2

Potassium

2.2

Copper

4.5

Sodium

2.3

Silver

4.7

Lithium

2.5

Platinum

5.6

Experimental set-up of PEE The experimental set-up to study the photoelectric effect is shown in fig. When monochromatic light of frequency greater than f0 is incident on the cathode, photoelectrons are emitted from it and they move towards anode A. Initially, the space between the cathode and the anode contains a number of electrons making up electron cloud. This negative charge repels the fresh electrons coming from the cathode. The electrons of maximum kinetic energy are able to reach the anode and constitutes a photocurrent. If anode is made positive with respect to the cathode, the emitted electrons are attracted by the anode and the photoelectric current increases. With the increase in anode potential, the photoelectric current increases and becomes maximum. Thereafter current will not increase with the increase in anode potential. This maximum value of current is called the saturation current (is). This will happen when all the emitted electrons by the cathode in any time interval are attracted by the anode . Fig. shows the photoelectric current i with anode potential V.

Stopping potential When anode is given negative potential with respect to the cathode, the photoelectric current decreases. For a particular value of anode potential, the photoelectric current becomes zero. The minimum negative anode potential at which photoelectric current becomes zero is called stopping or cut off potential V0. To stop the photoelectric current, we must ensure that even the fastest electron will not reach the anode. Thus stopping potential is related to the maximum kinetic energy of the ejected electrons. If V0 is the stopping potential, then Kmax = eV0.

...(4)

Characteristics of pee 1.

Effect of intensity of incident light When the intensity of the light increases, more number of photons strike with the photometal and thereby liberate more number of electrons. Because of this, photo current increases. As the frequency of the incident light is same, so maximum kinetic energy and hence slopping potential remains same.

2.

Effect of frequency of incident light When the intensity of incident light is kept constant and its frequency increases, the number of photons remains same but their kinetic energy increases. Therefore the emitted electrons are same in number but of greater kinetic energy and hence stopping potential also increases. Fig. shows the variation of photocurrent with frequency of light f. Buy books : http://www.dishapublication.com/entrance-exams-books/medical-exams.html

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3.

Effect of photometal When intensity and frequency of incident light are kept constant and photometal is changed, the stopping potential V0 versus frequency f are parallel straight lines. This shows that the slope V0/f is same for all metals and is equal to universal constant (h). If the graph is plotted between Kmax and f, then there is straight line. Slope of which gives the value of h/e (Fig.).

4.

Effect of time Metal starts emitting electrons as soon as light is incident on it and so there is no time lag between incident light and emitted electrons.

Einstein's explanation of pee Einstein forwarded the Plank¢s quantum theory to explain photoelectric effect. According to him light is made of small energy bundles, called photons. The energy of photon is proportional to the frequency f. That is E

=

hf,

where h is a universal constant, called Plank¢s constant. He made the following assumptions : 1.

The photoelectric effect is the result of collisions between photons of incident light and free electrons of the metal.

2.

The electrons of metal are bound with the nucleus by attractive forces. The minimum energy required to liberate an electron from this binding is called work function W0.

3.

The incident photon interacts with a single electron and spend energy in two parts : (i)

in liberating the electron from the metal surface,

(ii) and imparting kinetic energy to emitted electrons. Thus if hf is the energy of incident photons, then h f = W0 + Kmax ...(i) As

\

f hc l

=

hc c and W0 = l , l 0

=

hc 1 2 + mvmax . l0 2

...(ii)

Above equation is known as Einstein photo-electric equation. It should be remembered that photoelectric effect will occur only if l £ l o . 4.

The efficiency of photoelectric effect is less than 1%, i.e., only less than 1% of photons are capable of ejecting electrons from the metal surface. The rest 99% of the photon energy will convert into thermal energy.

5.

If V0 is the stopping potential, then Kmax = eV0 and so hf

=

W0 + eV0

or

hf e

=

W0 + V0 e

or

V0

=

-

W0 æ h ö + ç ÷ f. e èeø

...(iii)

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æhö The equation (iii) is a straight line between V0 and f, whose slope is ç ÷ , which èeø is a universal constant.