Atoms and Nuclei — Page 14 | JEE Physics

Q131

The mass of proton, neutron and helium nucleus are respectively

1.0073~u,1.0087~u

and

4.0015~u

. The binding energy of helium nucleus is :

A

28.4~\mathrm{MeV}

B

56.8~\mathrm{MeV}

C

7.1~\mathrm{MeV}

D

14.2~\mathrm{MeV}

Correct Answer

Option A

Solution

Mass defect $=2($ Mass of $p+$ mass of $n)-$ mass of $\mathrm{He}$ nucleus

\begin{aligned} & \Delta m=0.0305 u \\\\ & \text{ B.E }=931.5 \times \Delta m=931.5 \times 0.0305 \\\\ & =28.4 \mathrm{MeV} \end{aligned}

Q132

A free neutron decays into a proton but a free proton does not decay into neutron. This is because

A neutron is an uncharged particle

B neutron has larger rest mass than proton

C neutron is a composite particle made of a proton and an electron

D proton is a charged particle

Correct Answer

Option B

Solution

As neutron has more rest mass than proton it will require energy to decay proton into neutron.

Q133

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A: The nuclear density of nuclides

{ }_{5}^{10} \mathrm{~B},{ }_{3}^{6} \mathrm{Li},{ }_{26}^{56} \mathrm{Fe},{ }_{10}^{20} \mathrm{Ne}

and

{ }_{83}^{209} \mathrm{Bi}

can be arranged as

\rho_{\mathrm{Bi}}^{\mathrm{N}}>\rho_{\mathrm{Fe}}^{\mathrm{N}}>\rho_{\mathrm{Ne}}^{\mathrm{N}}>\rho_{\mathrm{B}}^{\mathrm{N}}>\rho_{\mathrm{Li}}^{\mathrm{N}}

Reason R: The radius

R

of nucleus is related to its mass number

A

as

R=R_{0} A^{1 / 3}

, where

R_{0}

is a constant. In the light of the above statements, choose the correct answer from the options given below

A

{Both ~\mathbf{A}}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

B Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

C

\mathbf{A}

is false but

\mathbf{R}

is true

D

\mathbf{A}

is true but

\mathbf{R}

is false

Correct Answer

Option C

Solution

R = {R_0}{A^{{1 \over 3}}}

, using this

\rho = {M \over {{4 \over 3}\pi {R^3}}} = {{A{m_P}} \over {{4 \over 3}\pi R_0^3A}} = {{{m_P}} \over {{4 \over 3}\pi R_0^3}}

$\rho$ is independent of mass number. $\therefore$ A is false

Q134

Speed of an electron in Bohr's

7^{\text{th }}

orbit for Hydrogen atom is

3.6 \times 10^{6} \mathrm{~m} / \mathrm{s}

. The corresponding speed of the electron in

3^{\text{rd }}

orbit, in

\mathrm{m} / \mathrm{s}

is :

A

\left(1.8 \times 10^{6}\right)

B

\left(7.5 \times 10^{6}\right)

C

\left(8.4 \times 10^{6}\right)

D

\left(3.6 \times 10^{6}\right)

Correct Answer

Option C

Solution

v\,\alpha {z \over n}

{{{v_1}} \over {{v_2}}} = \left( {{{{n_2}} \over {{n_1}}}} \right)

\Rightarrow {{3.6 \times {{10}^6}} \over {{v_2}}} = {3 \over 7}

\Rightarrow {v_2} = {7 \over 3} \times 3.6 \times {10^6}

m/s

= 8.4 \times {10^6}

m/s

Q135

Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of

\dfrac{1}{2}

day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days.

A

1.69\times10^{24}

B

3.38\times10^{24}

C

6.76\times10^{23}

D

6.76\times10^{24}

Correct Answer

Option B

Solution

{n_A} = 20

moles

{n_B} = 10

moles

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

{N_A} = (20\,N){e^{ - \left( {{{\ln 2} \over 1} \times 2} \right)}}

= {{20\,N} \over 4} = 5\,N

(N = Avogadro's Number)

{N_B} = 10\,N\,{e^{ - 4\ln 2}}

= \left( {{{10\,N} \over {16}}} \right)

{N_A} + {N_B} = 5\,N + {{10\,N} \over {16}} = \left( {{{90\,N} \over {16}}} \right) = 3.38 \times {10^{24}}

Q136

If a radioactive element having half-life of

30 \mathrm{~min}

is undergoing beta decay, the fraction of radioactive element remains undecayed after

90 \mathrm{~min}

. will be

A

\dfrac{1}{16}

B

\dfrac{1}{4}

C

\dfrac{1}{8}

D

\dfrac{1}{2}

Correct Answer

Option C

Solution

$t_{\text{half }}=30 \mathrm{~min}$ . In 90 min. there will be 3 half lives

\begin{aligned} \text{ Number of remaining } & =\left(\frac{N_{0}}{2^{3}}\right) \\\\ & =\frac{N_{0}}{8} \end{aligned}

$\therefore \quad$ Fraction will be $\dfrac{1}{8}$

Q137

The ratio of the density of oxygen nucleus (

_8^{16}O

) and helium nucleus (

_2^{4}\mathrm{He}

) is

A 4 : 1

B 1 : 1

C 2 : 1

D 8 : 1

Correct Answer

Option B

Solution

Nuclear density is independent of mass number As nuclear density $=\dfrac{\mathrm{Au}}{\dfrac{4}{3} \pi \mathrm{R}^3}$ Also, $\mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{\dfrac{1}{3}}$ And $R^3=R_0^3 A$ $\Rightarrow$ Nuclear density $=\dfrac{\mathrm{Au}}{\dfrac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}}$ Nuclear density $=\dfrac{3 \mathrm{u}}{4 \pi \mathrm{R}_0^3}$ $\Rightarrow$ Nuclear density is independent of $\mathrm{A}$

Q138

A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4

\times

10

^{-15}

eVs) :

A 99.3 nm

B 94.1 nm

C 974 nm

D 941 nm

Correct Answer

Option B

Solution

$\dfrac{h c}{\lambda}=+13.6 \mathrm{eV}\left[\dfrac{1}{1}-\dfrac{1}{4^{2}}\right]$

\Rightarrow \frac{4 \times 10^{-15} \times 3 \times 10^{-8}}{\lambda}=13.6\left[\frac{15}{16}\right]

$\lambda=94.1 \mathrm{~nm}$

Q139

If the wavelength of the first member of Lyman series of hydrogen is

\lambda

. The wavelength of the second member will be

A

\dfrac{27}{5} \lambda

B

\dfrac{5}{27} \lambda

C

\dfrac{27}{32} \lambda

D

\dfrac{32}{27} \lambda

Correct Answer

Option C

Solution

\begin{aligned} & \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\ & \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)} \end{aligned}

On dividing (i) & (ii)

\lambda^{\prime}=\frac{27}{32} \lambda

Q140

The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is:

A

\dfrac{1}{8}

B

\dfrac{3}{4}

C

\dfrac{7}{8}

D

\dfrac{1}{4}

Correct Answer

Option C

Solution

The decay of a radioactive nucleus is an exponential process, and the fraction of the original sample that remains after time $t$ is given by: $N(t) = N_0 e^{-\lambda t}$ where $N_0$ is the initial number of nuclei, $N(t)$ is the number of nuclei remaining after time $t$ , and $\lambda$ is the decay constant, which is related to the half-life $T_{1/2}$ by the equation: $\lambda = \dfrac{\ln(2)}{T_{1/2}}$ In this problem, the half-life of the nucleus is given as 5 years, so we have: $\lambda = \dfrac{\ln(2)}{5~\mathrm{yrs}} \approx 0.1386~\mathrm{yr^{-1}}$ We are asked to find the fraction of the original sample that would decay in 15 years.

At $t=15$ years, the fraction of nuclei remaining is: $N(15) = N_0 e^{-\lambda (15~\mathrm{yrs})}$ To find the fraction that has decayed, we subtract this expression from 1, since the fraction that remains plus the fraction that has decayed must add up to 1: Fraction decayed = $1 - N(15) = 1 - N_0 e^{-\lambda (15~\mathrm{yrs})}$ We know that the half-life of the nucleus is 5 years, which means that the fraction of nuclei remaining after one half-life is 1/2.

Therefore, after 3 half-lives (which is equivalent to 15 years), the fraction of nuclei remaining is: $N(15) = N_0 \left(\dfrac{1}{2}\right)^3 = \dfrac{N_0}{8}$ Plugging this into the equation for the fraction of nuclei that have decayed, we get: Fraction decayed = $1 - N_0 e^{-\lambda (15~\mathrm{yrs})} = 1 - \dfrac{N_0}{8} = \dfrac{7N_0}{8}$ Therefore, the fraction of the original sample that would decay in 15 years is $\dfrac{7}{8}$