Atoms and Nuclei — Page 5 | JEE Physics

Q41

This question contains Statement- 1 and Statement- 2. Of the four choices given after the statements, choose the one that best describes the two statements: Statement- 1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and Statement- 2: For heavy nuclei, binding energy per nucleon increases with increasing

Z

while for light nuclei it decreases with increasing

Z.

A Statement -

1

is false, Statement -

2

is true

B Statement -

1

is true, Statement -

2

is true; Statement -

2

is a correct explanation for Statement -

1

C Statement -

1

is true, Statement -

2

is true; Statement -

2

is not a correct explanation for Statement -

1

D Statement -

1

is true, Statement -

2

is false

Correct Answer

Option D

Solution

We know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement

(1)

is correct.

The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing

Z

and for light nuclei, B.E/nucleon increases with increasing

Z

and for light nuclei,

B.E/

nucleon increases with increasing

Z.

Q42

Suppose an electron is attracted towards the origin by a force

{k \over r}

where

'k'

is a constant and

'r'

is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the

{n^{th}}

orbital of the electron is found to be

'{r_n}'

and the kinetic energy of the electron to be

'{T_n}'.

Then which of the following is true?

A

{T_n} \propto {1 \over {{n^2}}},{r_n} \propto {n^2}

B

{T_n}

independent of

n,{r_n} \propto n

C

{T_n} \propto {1 \over n},{r_n} \propto n

D

{T_n} \propto {1 \over n},{r_n} \propto {n^2}

Correct Answer

Option B

Solution

When

F = {k \over r} =

centripetal force, then

{k \over r} = {{m{v^2}} \over r}

\Rightarrow m{v^2} =

constant $\Rightarrow$ kinetic energy is constant

\Rightarrow T

is independent of

n.

Q43

The transition from the state

n=4

to

n=3

in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from :

A

3 \to 2

B

4 \to 2

C

5 \to 4

D

2 \to 1

Correct Answer

Option C

Solution

It is given that transition from the state

n=4

to

n=3

in a hydrogen like atom result in ultraviolet radiation.

For infrared radiation the energy gap should be less.

The only option is

5

\to 4.

Q44

A nucleus of mass

M+

\Delta m

is at rest and decays into two daughter nuclei of equal mass

{M \over 2}

each. Speed of light is

c.

The speed of daughter nuclei is

A

c{{\Delta m} \over {M + \Delta m}}

B

c\sqrt {{{2\Delta m} \over M}}

C

c\sqrt {{{\Delta m} \over M}}

D

c\sqrt {{{\Delta m} \over {M + \Delta m}}}

Correct Answer

Option B

Solution

By conservation of energy,

\left( {M + \Delta m} \right){c^2} = {{2M} \over 2}{c^2} + {1 \over 2}.{{2M} \over 2}{v^2},

where

v

is the speed of the daughter nuclei

\Rightarrow \Delta m{c^2} = {M \over 2}{v^2}

$\therefore$

v = c\sqrt {{{2\Delta m} \over M}}

Q45

The half life of a radioactive substance is

20

minutes. The approximate time interval

\left( {{t_2} - {t_1}} \right)

between the time

{{t_2}}

when

{2 \over 3}

of it had decayed and time

{{t_1}}

when

{1 \over 3}

of it had decayed is :

A

14

min

B

20

min

C

28

min

D

7

min

Correct Answer

Option B

Solution

Number of undecayed atom after time

{t_2};

{{{N_0}} \over 3} = {N_0}{e^{ - \lambda {t_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

Number of undecayed atom after time

{t_1};

{{2{N_0}} \over 3} = {N_0}{e^{ - \lambda {t_1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

From

(i),

{e^{ - \lambda {t_2}}} = {1 \over 3}

\Rightarrow - \lambda {t_2} = {\log _e}\left( {{1 \over 3}} \right)\,\,\,\,\,\,\,\,...\left( {iii} \right)

From

(ii)

- {e^{ - \lambda {t_2}}} = {2 \over 3}

\Rightarrow - \lambda {t_1} = {\log _e}\left( {{2 \over 3}} \right)\,\,\,\,\,\,\,\,\,\,...\left( {iv} \right)

Solving

(iii)

and

(iv),

we get

{t_2} - {t_1} = 20\,

min

Q46

Energy required for the electron excitation in

L{i^{ + + }}

from the first to the third Bohr orbit is :

A

36.3

eV

B

108.8

eV

C

122.4

eV

D

12.1

eV

Correct Answer

Option B

Solution

Energy of excitation,

\Delta E = 13.6\,{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)eV

\Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right) = 108.8\,eV

Q47

Hydrogen atom is excited from ground state to another state with principal quantum number equal to

4.

Then the number of spectral lines in the emission spectra will be :

A

2

B

3

C

5

D

6

Correct Answer

Option D

Solution

The possible number of the spectral lines is given

= {{n\left( {n - 1} \right)} \over 2} = {{4\left( {4 - 1} \right)} \over 2} = 6

Q48

Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron

= 1.6725 \times {10^{ - 27}}kg,

mass of proton

= 1.6725 \times {10^{ - 27}}\,kg,

mass of electron

= 9 \times {10^{ - 31}}\,kg

).

A

0.51

MeV

B

7.10\,MeV

C

6.30\,MeV

D

5.4\,MeV

Correct Answer

Option A

Solution

{}_0^1n \to {}_1^1H + {}_{ - 1}{e^0} + \overrightarrow v + Q

The mass defect during the process

\Delta m = {m_n} - {m_H} - {m_e}

= 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right)

= - 9 \times {10^{ - 31}}kg

The energy released during the process

E = \Delta m{c^2}

E = 9 \times {10^{ - 31}} \times 9 \times {10^{16}}

= 81 \times {10^{ - 15}}\,joules

E = {{81 \times {{10}^{ - 15}}} \over {1.6 \times {{10}^{ - 19}}}} = 0.511MeV

Q49

A diatomic molecule is made of two masses

{m_1}

and

{m_2}

which are separated by a distance

r.

If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: (

n

is an integer)

A

{{{{\left( {{m_1} + {m_2}} \right)}^2}{n^2}{h^2}} \over {2m_1^2m_2^2{r^2}}}

B

{{{n^2}{h^2}} \over {2\left( {{m_1} + {m_2}} \right){r^2}}}

C

{{2{n^2}{h^2}} \over {\left( {{m_1} + {m_2}} \right){r^2}}}

D

{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}} \over {2{m_1}{m_2}{r^2}}}

Correct Answer

Option D

Solution

The energy of the system of two atoms of diatomic molecule

E = {1 \over 2}I\omega

where

I=

moment of inertia

\omega =

Angular velocity

= {L \over I}.

L=

Angular momentum

I = {1 \over 2}\left( {{m_1}{r_1}^2 + {m^2}{r_2}^2} \right)

Thus,

E = {1 \over 2}\left( {{m_1}{r_1}^2 + m{}_2{r_2}^2} \right){\omega ^2}\,\,\,\,\,\,\,\,\,...\left( i \right)

E = {1 \over 2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right){{{L^2}} \over {{I^2}}}

L = n{{nh} \over {2n}}

(According Bohr's Hypothesis)

E = {1 \over 2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right){{{L^2}} \over {{{\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}^2}}}

E = {1 \over 2}{{{L^2}} \over {\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}}

= {{{n^2}{h^2}} \over {8{\pi ^2}\left( {{m_1}{r_1}^2 + {m_2}{r_2}^2} \right)}}

E = {{\left( {{m_1} + {m_2}} \right){n^2}{h^2}} \over {8{\pi ^2}{r^2}{m_1}{m_2}}}

\left[ {\,\,} \right.

as

\left. {\,\,\,{r_1} = {{{m_2}r} \over {{m_1} + {m_2}}};\,{r_2} = {{{m_2}r} \over {{m_1} + {m_2}}}\,\,} \right]

Q50

The radiation corresponding to

3 \to 2

transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field

3 \times {10^{ - 4}}\,T.

If the radius of the larger circular path followed by these electrons is

10.0

mm

, the work function of the metal is close to:

A

1.8

eV

B

1.1

eV

C

0.8

eV

D

1.6

eV

Correct Answer

Option B

Solution

Radius of circular path followed by electron is given by,

r = {{m\upsilon } \over {qB}} = {{\sqrt {2meV} } \over {eB}} = {1 \over B}\sqrt {{{2m} \over e}V}

\Rightarrow V = {{{B^2}{r^2}e} \over {2m}} = 0.8V

For transition between

3

to

2.

E = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)

= {{13.6 \times 5} \over {36}} = 1.88eV

Work function

= 1.88eV - 0.8eV

= 1.08eV \approx 1.1eV