Atoms And Nuclei(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

ATOMS AND NUCLEI MCQs

Atoms, Nuclei (12th Grade)

Total Questions : 60 | Page 5 of 6 pages

Question 41. Radius of first orbit of hydrogen is

0.53 ˙ A

. The radius in fourth orbit is:

8.48˙A
2.12˙A
4.24˙A
0.193˙A

Discuss Question

Answer: Option A. -> 8.48˙A
:
A

r_{n} \propto n^{2}

\frac{r_{4}}{r_{1}} = {[\frac{4}{1}]}^{2} = 16

r_{4} = 16 \times 0.53

= 8.48 ˙ A

Question 42. A hydrogen atom in the ground state emits a photon of energy 12.1 eV. Its orbital angular momentum changes by

Δ

L. Then

Δ

L is equal to

1.05×10−34 Js
2.11×10−34 Js
3.16×10−34 Js
4.22×10−34 Js

Discuss Question

Answer: Option B. -> 2.11×10−34 Js
:
B

Δ E = 12.1 = E_{1} - E_{n}

E_{n} = 13.6 e V - 12.1

E_{n} = 1.5

\Rightarrow n = 3

L_{1} = \frac{2}{2 π}

L_{3} = \frac{3 h}{2 π}

Δ L = L_{3} - L_{1} = \frac{3 h}{2 π} - \frac{h}{2 π} = \frac{2 h}{2 π} = \frac{h}{π} = \frac{6.6}{3.14} \times 10^{- 34} s^{- 1}

Δ L = 2.11 \times 0^{- 34} J s

Question 43. When a silver foil (Z = 47) was used in a

α

-ray scattering experiment, the number of

α

particles scattered at

30^{\circ}

was found to be 200 per minute. If the silver foil is replaced by aluminium (z = 13) foil of same thickness, the number of

α

-particles scattered per minute at

30^{\circ}

is nearly equal to

Discuss Question

Answer: Option A. -> 15
:
A

N \propto Z^{2}

\frac{N_{2}}{N_{1}} = {[\frac{z_{2}}{z_{1}}]}^{2} = {[\frac{13}{47}]}^{2}

\frac{N_{2}}{200} = \frac{13 \times 13}{47 \times 47}

N_{2} = \frac{13 \times 13 \times 200}{47 \times 47} = 15

Question 44. An electron in a hydrogen atom makes a transition

n_{1} \to n_{2}

where

n_{1}

and

n_{2}

are principal quantum numbers of the states. Assume the Bhor's model to be valid. The time period of the electron in the initial states is eight times to that of final state. What is ratio of

\frac{n_{1}}{n_{2}}

8 : 1
4 : 1
2 : 1
1 : 2

Discuss Question

Answer: Option C. -> 2 : 1
:
C

T_{n} \propto n^{3}

n \propto T^{\frac{1}{3}}

\frac{n_{1}}{n_{2}} = {[\frac{T_{1}}{T_{2}}]}^{\frac{1}{3}} = {[\frac{8}{1}]}^{\frac{1}{3}}

n_{1} : n_{2} = 2 : 1

Question 45. Let

ϑ_{1}

be the frequency of the series limit of the Lyman series and

ϑ_{2}

be the frequency of the first line of the Lyman series

ϑ_{3}

be the frequency of the series limit of Balmar series, then

ϑ1−ϑ2=ϑ3
ϑ2−ϑ1=ϑ3
2ϑ3=ϑ1+ϑ2
ϑ1=ϑ2−ϑ3

Discuss Question

Answer: Option A. -> ϑ1−ϑ2=ϑ3
:
A

E_{\infty} - E_{1} = h ν_{1} . . . . (1)

E_{\infty} - E_{2} = h ν_{3} . . . . (2)

E_{2} - E_{1} = h ν_{2} . . . . (3)

(1) - (2)

E_{2} - E_{1} = h [ν_{1} - ν_{3}] . . . . (4)

From (3) and (4)

h ν_{2} = h [ν_{1} - ν_{3}]

ν_{3} = ν_{1} - ν_{2}

Question 46. In the

n^{t h}

orbit, the energy of an electron

E_{n} = - \frac{13.6}{n^{2}} e V

for hydrogen atom. The energy required to take the electron from first orbit to second orbit will be

10.2 eV
12.1 eV
13.6 eV
3.4 eV

Discuss Question

Answer: Option A. -> 10.2 eV
:
A
The energy in the first orbit is - 13.6eV

n = 2

---------------

E_{2} = - \frac{13.6}{(2)^{2}} = - 3.4 e V

E_{1 \to 2} = - 3.4 - (13.6) = + 10.2 e V

Question 47. The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is

20, 397 c m^{- 1}

. The wave number of the energy for the same transition in

H e^{+}

5,099 cm−1
20,497 cm−1
40,994 cm−1
81,588 cm−1

Discuss Question

Answer: Option D. -> 81,588 cm−1
:
D
Using

\frac{1}{λ} = ¯ v = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \Rightarrow ¯ v \infty Z^{2} \Rightarrow \frac{_{2}}{_{1}} = {(\frac{Z_{2}}{Z_{1}})}^{2} = {(\frac{Z}{1})}^{2} = 4 \Rightarrow_{2} = ¯ v \times 4 = 81588 c m^{- 1}

Question 48. In an atom, the two electrons move round the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is

Discuss Question

Answer: Option D. -> 18
:
D
Time period

T \propto \frac{n^{3}}{Z^{2}}

For a given atom (Z = constant) So

T \propto n^{3}

......... (i) and radius

R \propto n^{2}

......... (ii)

∴

From equation (i) and (ii)

T \propto R^{3 / 2} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{R_{1}}{R_{2}})}^{3 / 2} = {(\frac{R}{4 R})}^{3 / 2} = \frac{1}{8}

Question 49. Hydrogen atom in its ground state is excited by radiation of wavelength

975 \circ A

. How many lines will be there in
the emission spectrum

Discuss Question

Answer: Option C. -> 6
:
C
Using

\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \Rightarrow \frac{1}{975 \times 10^{- 10}} = 1.097 \times 10^{7} (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow n = 4

Now number of spectral lines

N = \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6

Question 50. In the following atoms and molecules for the transition from n = 2 to n = 1, the spectral line of minimum wavelength will be produced by