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Vidyamandir Classes

Solutions to Home Work Test/Chemistry Atomic Structure

HWT - 1

1.(C)

In lightest nuclei there is only one proton and zero neutrons and mp (1876)me

4.(A)

Is and 25 are considered degenerate orbitals.

5.(A)

Number of photoelectrons is proportional to intensity if incident beam only if frequency of light is greater than threshold frequency.

7.(B)

K . E max h h0 K.E max 6.63 1019 J 1015 Hz h 6.63 1034 J H 2 1

0 8.

For Bohr’s theory, only one proton should be resent.

Atomic Structure 1.(C)

0 (2 1) 1015 Hz 1015 Hz

HWT - 2

1 1 E 13.6 ev 2 (n 2 )2 (n1 ) Put n 2 And n1 1

2.

Fact

3.

K.E. eV

[V : Change in potential]

1 m2 eV 2

(e : charge]

V

2eV m

h h h p mV 2m(K.E)

4.

5.

In n = 4 we have 16 orbitals

8.

Calcium has z = 20

10.

1s2 2s2 2p6 3s2 3p6 4s2

z = 28

for nickel

Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8

1 K.E

Ni 2 1s2 2s2 2p6 3s2 3p6 4s0 3d8 3d8

2 unpaired e s

VMC/Atomic Structure

23

HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure 1. 2.

HWT - 3

h mV Shortest wavelength is given when largest energy change is there. for H it is n = 1 to n Use

And for longest wavelength, look for smallest energy charge

He2 n = 3 to n = 2. Let wavelength by y.

1 1 1 R (1)2 2 R x ()2 (1) 1 R (2)2 y

x 5 y 9

8.

1 1 (5) 2 2 R 4 49 (3) (2)

y

9x 5

Radial nodes are given by : n 1

Atomic Structure

HWT - 4

1.

Hydrogen atom has 1 proton and 1 electron. ionized hydrogen atom has only 1 proton.

5.

r

n2 z

6.

Angular Momentum =

8.

In module

9.

In module

nh h n : 2 2

Atomic Structure 1.

HWT - 5

k(z = 19) we have 1s2 2s2 2p2 3s2 3p6 4s1 Outermost e is 4s1 1 1 4, 0, 0, + or 4, 0, 0, . 2 2 h Use [p mv] 5. p

2.

T

1 n

6.

Splitting of spectral lines in electric field is Stark effect. Splitting of spectral lines in magnetic field is leeman effect.

7.

Fact

9.

Isobars have equal atomic mass [protons + neutrons] Istotoes have equal atomic no. [protons] Isotones have equal neutrons.

Atomic Structure

HWT - 6 nh 2

1.

for d orbital 2

4.

Angular momentum

5.

Fact

6.

Fact

7.

When ‘n + e’ values lashes we look for value of n and smaller the nature of n larger is the stability.

8.

Isotones have equal neutrons.

VMC/Atomic Structure

10.

1 1 Use E 2 (n 2 )2 (n1 )

24

HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure

HWT – 7

1.

Hund’s rule is niolated in 1st case.

6.

Use K . E max hv w 0

7.

Spherical nodes = n 1

9.

Use Aufbav’s principle to know the configuration

Atomic Structure 1.

4.

6.

HWT - 8

These transmissions belong to lyman series Uv rays

2 unpaired e s

Ni 2 has 3d8 Energy of photon

1

hc as E

Atomic Structure 2.

Add all e s to get atomic number here

3.

z = 21

HWT - 9

1s2 2s2 2p6 3s2 3p6 4s2 3d1 4.

12 ‘p’ electrons

5

8.

Fact

10.

1 ‘d’ electon. h h Use p mv 3d1

Atomic Structure 1.

HWT - 10

Cl (z = 17) 1s2 2s2 2p6 3s2 3p5

n=3 e=1 m could be l or.

4.

Fact.

5.

3s, 3p, 3d 1 + 3 + 5 = 9 orbitals

9.

Definition of Hund’s rule.

VMC/Atomic Structure

25

HWT-Solutions/Chemistry

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Solutions to Home Work Test/Chemistry Atomic Structure

HWT - 1

1.(C)

In lightest nuclei there is only one proton and zero neutrons and mp (1876)me

4.(A)

Is and 25 are considered degenerate orbitals.

5.(A)

Number of photoelectrons is proportional to intensity if incident beam only if frequency of light is greater than threshold frequency.

7.(B)

K . E max h h0 K.E max 6.63 1019 J 1015 Hz h 6.63 1034 J H 2 1

0 8.

For Bohr’s theory, only one proton should be resent.

Atomic Structure 1.(C)

0 (2 1) 1015 Hz 1015 Hz

HWT - 2

1 1 E 13.6 ev 2 (n 2 )2 (n1 ) Put n 2 And n1 1

2.

Fact

3.

K.E. eV

[V : Change in potential]

1 m2 eV 2

(e : charge]

V

2eV m

h h h p mV 2m(K.E)

4.

5.

In n = 4 we have 16 orbitals

8.

Calcium has z = 20

10.

1s2 2s2 2p6 3s2 3p6 4s2

z = 28

for nickel

Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8

1 K.E

Ni 2 1s2 2s2 2p6 3s2 3p6 4s0 3d8 3d8

2 unpaired e s

VMC/Atomic Structure

23

HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure 1. 2.

HWT - 3

h mV Shortest wavelength is given when largest energy change is there. for H it is n = 1 to n Use

And for longest wavelength, look for smallest energy charge

He2 n = 3 to n = 2. Let wavelength by y.

1 1 1 R (1)2 2 R x ()2 (1) 1 R (2)2 y

x 5 y 9

8.

1 1 (5) 2 2 R 4 49 (3) (2)

y

9x 5

Radial nodes are given by : n 1

Atomic Structure

HWT - 4

1.

Hydrogen atom has 1 proton and 1 electron. ionized hydrogen atom has only 1 proton.

5.

r

n2 z

6.

Angular Momentum =

8.

In module

9.

In module

nh h n : 2 2

Atomic Structure 1.

HWT - 5

k(z = 19) we have 1s2 2s2 2p2 3s2 3p6 4s1 Outermost e is 4s1 1 1 4, 0, 0, + or 4, 0, 0, . 2 2 h Use [p mv] 5. p

2.

T

1 n

6.

Splitting of spectral lines in electric field is Stark effect. Splitting of spectral lines in magnetic field is leeman effect.

7.

Fact

9.

Isobars have equal atomic mass [protons + neutrons] Istotoes have equal atomic no. [protons] Isotones have equal neutrons.

Atomic Structure

HWT - 6 nh 2

1.

for d orbital 2

4.

Angular momentum

5.

Fact

6.

Fact

7.

When ‘n + e’ values lashes we look for value of n and smaller the nature of n larger is the stability.

8.

Isotones have equal neutrons.

VMC/Atomic Structure

10.

1 1 Use E 2 (n 2 )2 (n1 )

24

HWT-Solutions/Chemistry

Vidyamandir Classes Atomic Structure

HWT – 7

1.

Hund’s rule is niolated in 1st case.

6.

Use K . E max hv w 0

7.

Spherical nodes = n 1

9.

Use Aufbav’s principle to know the configuration

Atomic Structure 1.

4.

6.

HWT - 8

These transmissions belong to lyman series Uv rays

2 unpaired e s

Ni 2 has 3d8 Energy of photon

1

hc as E

Atomic Structure 2.

Add all e s to get atomic number here

3.

z = 21

HWT - 9

1s2 2s2 2p6 3s2 3p6 4s2 3d1 4.

12 ‘p’ electrons

5

8.

Fact

10.

1 ‘d’ electon. h h Use p mv 3d1

Atomic Structure 1.

HWT - 10

Cl (z = 17) 1s2 2s2 2p6 3s2 3p5

n=3 e=1 m could be l or.

4.

Fact.

5.

3s, 3p, 3d 1 + 3 + 5 = 9 orbitals

9.

Definition of Hund’s rule.

VMC/Atomic Structure

25

HWT-Solutions/Chemistry

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