Atoms and Nuclei — Page 11 | JEE Physics

Q101

Consider the following statements : A. Atoms of each element emit characteristics spectrum. B. According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit. C. The density of nuclear matter depends on the size of the nucleus. D. A free neutron is stable but a free proton decay is possible. E. Radioactivity is an indication of the instability of nuclei. Choose the correct answer from the options given below :

A A, B, C, D and E

B A, B and E only

C B and D only

D A, C and E only

Correct Answer

Option B

Solution

(A) True, atom of each element emits characteristic spectrum. (B) True, according to Bohr's postulates

mvr = {{nh} \over {2\pi }}

and hence electron resides into orbits of specific radius called stationary orbits.

(C) False, density of nucleus is constant.

(D) False, A free neutron is unstable decays into proton and electron and antineutrino.

(E) True, unstable nucleus show radioactivity.

Q102

A particular hydrogen like ion emits radiation of frequency 2.92

\times

1015 Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be :

A 0.44

\times

1015

B 6.57

\times

1015

C 4.38

\times

1015

D 2.46

\times

1015

Correct Answer

Option D

Solution

n{f_1} = k\left( {{1 \over 1} - {1 \over {{3^2}}}} \right)

n{f_2} = k\left( {1 - {1 \over {{2^2}}}} \right)

{{{f_1}} \over {{f_2}}} = {{8/9} \over {3/4}} \Rightarrow {f_2} = 2.46 \times {10^{15}}

Q103

There are 1010 radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds?

\left( {\sqrt 2 = 1.414} \right)

A 2

\times

1010

B 7

\times

109

C 105

D 4

\times

1010

Correct Answer

Option B

Solution

{N \over {{N_0}}} = {\left( {{1 \over 2}} \right)^{{t \over {{t^{1/2}}}}}}

{N \over {{{10}^{10}}}} = {\left( {{1 \over 2}} \right)^{{{30} \over {60}}}}

\Rightarrow N = {10^{10}} \times {\left( {{1 \over 2}} \right)^{{1 \over 2}}} = {{{{10}^{10}}} \over {\sqrt 2 }} \approx 7 \times {10^9}

Q104

The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then :

A x-will decay faster than y.

B y-will decay faster than x.

C x and y have same decay rate initially and later on different decay rate.

D x and y decay at the same rate always.

Correct Answer

Option B

Solution

Given, ( $\tau$ 1/2)x = ( $\tau$ )y Here, $\tau$ 1/2 = half-life period of radioactive element and $\tau$ = mean life period of radioactive element.

As we know the expression, Half-life of the radioactive element x,

{\tau _{1/2}} = {{\ln (2)} \over {{\lambda _x}}}

Mean life of the radioactive element y,

\tau = {1 \over {{\lambda _y}}}

Substituting the values in Eq. (i), we get

{{\ln 2} \over {{\lambda _x}}} = {1 \over {{\lambda _y}}} \Rightarrow {\lambda _x} = 0.693{\lambda _y}

Initially they have same number of atoms,

{N_x} = {N_y} = {N_0}

As we know that, Activity,

A = \lambda N

As,

{\lambda _x} < {\lambda _y} \Rightarrow {A_x} < {A_y}

Therefore, y will decay faster than x.

Q105

The activity of a radioactive material is 2.56

\times

10

-

3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2

\times

10

-

5 Ci ?

A 30 days

B 35 days

C 40 days

D 25 days

Correct Answer

Option B

Solution

By Radioactive Decay law,

\begin{aligned} & \mathrm{R}=\mathrm{R}_{\mathrm{o}} e^{-\lambda t} \\\\ & \Rightarrow 2 \times 10^{-5}=2.56 \times 10^{-3} e^{-\lambda t} \end{aligned}

[Where, $\mathrm{R} =$ Activity at time $t$ $\lambda =$ Activity constant of Radioactive sample and

\lambda = \frac{\ln 2}{\mathrm{~T}_{1 / 2}}

Taking logarithm on both sides

\ln \left(2 \times 10^{-5}\right)=\ln \left(2.56 \times 10^{-3}\right)+\ln \left(e^{-\lambda t}\right)

\Rightarrow\ln \left(2 \times 10^{-5}\right)-\ln \left(2.56 \times 10^{-3}\right)=-\lambda t

$\Rightarrow \ln \left(\dfrac{2 \times 10^{-5}}{2.56 \times 10^{-3}}\right)=-\lambda t$ $\Rightarrow \ln \left(\dfrac{1}{128}\right)=-\lambda t$ $\Rightarrow-\ln 128=-\lambda t$ $\Rightarrow \ln 2^7=\dfrac{\ln 2}{\mathrm{~T}_{1 / 2}} t$ $\Rightarrow 7 \ln 2=\dfrac{\ln 2}{\mathrm{~T}_{1 / 2}} t$ $\Rightarrow t=7 \mathrm{~T}_{1 / 2}$ $\Rightarrow t=7 \times 5=35$ days

Q106

In the following nuclear reaction,

D\overset{\alpha}\longrightarrow {D_1}\overset{{{\beta ^ - }}}\longrightarrow {D_2}\overset{\alpha}\longrightarrow {D_3}\overset{\gamma}\longrightarrow {D_4}

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D4 respectively will be _________.

A 174 and 71

B 174 and 69

C 172 and 69

D 172 and 71

Correct Answer

Option A

Solution

Equivalent reaction can be written as

D\overset{{}}\longrightarrow {D_4} + 2\alpha + {\beta ^ - } + \gamma

$\Rightarrow$ Mass number of D4 = Mass number of D $-$ 2 $\times$ 4 = 182 $-$ 8 = 174 $\Rightarrow$ Atomic number of D4 = Atomic number of D $-$ 2 $\times$ 2 + 1 = 74 $-$ 4 + 1 = 71

Q107

Following statements related to radioactivity are given below : (A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions. (B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time. (C) Slope of the graph of loge (no. of undecayed nuclei) Vs. time represents the reciprocal of mean life time (

\tau

). (D) Product of decay constant (

\lambda

) and half-life time (T1/2) is not constant. Choose the most appropriate answer from the options given below :

A (A) and (B) only

B (B) and (D) only

C (B) and (C) only

D (C) and (D) only

Correct Answer

Option C

Solution

Radioactive decay is a random and spontaneous process it depends on unbalancing of nucleus.

N = {N_0}{e^{ - \lambda t}}

..... (B)

\ln N = - \lambda t + \ln {N_0}

So, slope

= - \lambda

..... (C)

{t_{1/2}} = {{\ln 2} \over \lambda }

So

{t_{1/2}} \times \lambda = \ln 2 =

Constant

Q108

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, Kp are related as :

A Q = Kp

B (Kp + Q) < 0

C Q p

D (Kp + Q) > 0

Correct Answer

Option D

Solution

Kp > 0 If Q is released $\Rightarrow$ Q > 0 $\Rightarrow$ Kp + Q > 0 Even the particle has to be given kinetic energy greater than magnitude of Q to maintain momentum conservation.

$\Rightarrow$ K + Q > 0

Q109

Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)

A 30.2 MeV

B 32.4 MeV

C 23.6 MeV

D 25.8 MeV

Correct Answer

Option C

Solution

1H2 + 1H2 $\to$ 2He4 No. of proton in one dueteron = 2

\therefore\,\,\,

Total protons in two dueterons = 2 $\times$ 2 = 4

\therefore\,\,\,

Binding energy of two dueteron = 1.1 $\times$ 4 = 4 : 4 MeV In (2He4) no of protons = 4

\therefore\,\,\,

Binding energy of (2He4) nuclei = 4 $\times$ 7 = 28 Mev Energy released in this process = 28 $-$ 4.4 = 23.6 MeV

Q110

Given below are two statements : Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1) to higher energy orbit (E2), is given as hf = E1

-

E2 Statement II : The jumping of electron from higher energy orbit (E2) to lower energy orbit (E1) is associated with frequency of radiation given as f = (E2

-

E1)/h This condition is Bohr's frequency condition. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true.

B Both Statement I and Statement II are false.

C Statement I is correct but Statement II is false.

D Statement I is incorrect but Statement II is true.

Correct Answer

Option D

Solution

Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.

Also, E2 $-$ E1 is the energy of photon $\Rightarrow$ E2 $-$ E1 = hf

\Rightarrow f = {{{E_2} - {E_1}} \over h}