Atoms and Nuclei — Page 3 | JEE Physics

Q21

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 binding energy per nucleon for the parent nucleus is

{E_1}

and that for the daughter nuclei is

{E_2}.

Then

A

{E_2} = 2{E_1}

B

{E_1} > {E_2}

C

{E_2} > {E_1}

D

{E_1} = 2{E_2}

Correct Answer

Option C

Solution

In nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleon of daughter nuclei is greater than the parent nucleus.

Q22

A radioactive nucleus (initial mass number

A

and atomic number

Z

emits

3\,\alpha

- particles and

2

positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

A

{{A - Z - 8} \over {Z - 4}}

B

{{A - Z - 4} \over {Z - 8}}

C

{{A - Z - 12} \over {Z - 4}}

D

{{A - Z - 4} \over {Z - 2}}

Correct Answer

Option B

Solution

{}_Z^AX\mathop \to \limits^{A - 12} {}_{Z - 8}Y + 3{}_2^4{X_e} + {}_t^0e

Number of protons, Np = Z - 8 Number of neutrons, Nn =

A - 12 - \left( {Z - 8} \right)

$\therefore$ Required ratio

= {{A - Z - 4} \over {Z - 8}}

Q23

In the nuclear fusion reaction

{}_1^2H + {}_1^3H \to {}_2^4He + n

given that the repulsive potential energy between the two nuclei is

\sim 7.7 \times {10^{ - 14}}J

, the temperature at which the gases must be heated to initiate the reaction is nearly [ Boltzmann's Constant

k = 1.38 \times {10^{ - 23}}\,J/K

$ ]

A

{10^7}\,\,K

B

{10^5}\,\,K

C

{10^3}\,\,K

D

{10^9}\,\,K

Correct Answer

Option D

Solution

The average kinetic energy per molecule

= {3 \over 2}kT

This kinetic energy should be able to provide the repulsive potential energy $\therefore$

{3 \over 2}kT = 7.7 \times {10^{ - 14}}

\Rightarrow T = {{2 \times 7.7 \times {{10}^{ - 14}}} \over {3 \times 1.38 \times {{10}^{ - 23}}}} = 3.7 \times {10^9}

Q24

If

13.6

eV

energy is required to ionize the hydrogen atom, then the energy required to remove an electron from

n=2

is

A

10.2

eV

B

0

eV

C

3.4

eV

D

6.8

eV.

Correct Answer

Option C

Solution

KEY CONCEPT : The energy of nth orbit of hydrogen is given by

{E_n} = {{13.6} \over {{n^2}}}eV/

atom For

n=2,

{E_n} = {{ - 13.6} \over 4} = - 3.4eV

Therefore the energy required to remove electron from

n = 2

is

+3.4eV.

Q25

A nucleus with

Z=92

emits the following in a sequence:

\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,{\beta ^ + },{\beta ^ + },\alpha

Then

Z

$ of the resulting nucleus is

A

76

B

78

C

82

D

74

Correct Answer

Option B

Solution

The number of $\alpha$ - particles released

=8

Therefore the atomic number should decrease by

16

The number of

{\beta ^ - }

- particles released

=4

Therefore the atomic number should increase by

4.

Also the number of

{\beta ^ + }

particles released is

2,

which should decrease the atomic number by

2.

Therefore the final atomic number is

92-16+4-2=78

Q26

When a

{U^{238}}

nucleus originally at rest, decays by emitting an alpha particle having a speed

'u',

the recoil speed of the residual nucleus is

A

{{4\mu } \over {238}}

B

- {{4\mu } \over {234}}

C

{{4\mu } \over {234}}

D

- {{4\mu } \over {238}}

Correct Answer

Option C

Solution

Here, conservation of linear momentum can be applied

238 \times 0 = 4u + 234v

$\therefore$

v = - {4 \over {234}}u

$\therefore$ speed

= |\overrightarrow v | = {4 \over {234}}u

Q27

The wavelengths involved in the spectrum of deuterium

\left( {{}_1^2\,D} \right)

are slightly different from that of hydrogen spectrum, because

A the size of the two nuclei are different

B the nuclear forces are different in the two cases

C the masses of the two nuclei are different

D the attraction between the electron and the nucleus is different in the two cases

Correct Answer

Option C

Solution

The wavelength of spectrum is given by

{1 \over \lambda } = R{z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

where

R = {{1.097 \times {{10}^7}} \over {1 + {m \over M}}}

where

m=

mass of electron

M=

mass of nucleus. For different

M,R

is different and therefore $\lambda$ is different

Q28

The binding energy per nucleon of deuteron

\left( {{}_1^2\,H} \right)

and helium nucleus

\left( {{}_2^4\,He} \right)

is

1.1

MeV

and

7

MeV

respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

A

23.6\,\,MeV

B

26.9\,\,MeV

C

13.9\,\,MeV

D

19.2\,\,MeV

Correct Answer

Option A

Solution

The nuclear reaction of process is

2_1^2H \to {4 \over 2}

He Energy released

= 4 \times \left( 7 \right) - 4\left( {1.1} \right) = 23.6\,MeV

Q29

A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of

2:1.

The ratio of their nuclear sizes will be

A

{3^{{{1}/{2}}}}:1

B

1:{2^{1/3}}

C

{2^{1/3}}:1

D

1:{3^{{{1}/{2}}}}

Correct Answer

Option B

Solution

From conservation of momentum

{m_1}{v_1} = {m_2}{v_2}

\Rightarrow \left( {{{{m_1}} \over {{m_2}}}} \right) = \left( {{{{v_2}} \over {{v_1}}}} \right)\,\,

given

\,\,{{{v_1}} \over {v{}_2}} = 2

\Rightarrow {{{m_1}} \over {{m_2}}} = {1 \over 2}

\Rightarrow {{r_1^3} \over {r_2^3}} = {1 \over 2}

\Rightarrow \left( {{{{r_1}} \over {{r_2}}}} \right) = {\left( {{1 \over 2}} \right)^{1/3}}

Q30

An

\alpha

-particle of energy

5

MeV

is scattered through

{180^ \circ }

by a fixed uranium nucleus. The distance of closest approach is of the order of

A

{10^{ - 12}}\,cm

B

{10^{ - 10}}\,cm

C

1A

D

{10^{ - 15a}}\,cm

Correct Answer

Option A

Solution

KEY NOTE : Distance of closest approach

{r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}

Energy,

E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J

$\therefore$

{r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)} \over {5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}

\Rightarrow r = 5.2 \times {10^{ - 14}}m = 5.3 \times {10^{ - 12}}cm