Atoms and Nuclei — Page 13 | JEE Physics

Q121

A nucleus of mass

M

at rest splits into two parts having masses

\dfrac{M^{\prime}}{3}

and $${{2M'} \over 3}(M'

A 1 : 2

B 2 : 1

C 1 : 1

D 2 : 3

Correct Answer

Option C

Solution

Linear momentum is conserved so,

{p_{M'/3}} = {p_{2M'/3}}

so,

{{{\lambda _{M'/3}}} \over {{\lambda _{2M'/3}}}} = {1 \over 1}

Q122

Mass numbers of two nuclei are in the ratio of

4: 3

. Their nuclear densities will be in the ratio of

A 4 : 3

B

\left(\dfrac{3}{4}\right)^{\dfrac{1}{3}}

C 1 : 1

D

\left(\dfrac{4}{3}\right)^{\dfrac{1}{3}}

Correct Answer

Option C

Solution

$\therefore$

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

\Rightarrow {{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{{1 \over 3}}} = {\left( {{4 \over 3}} \right)^{{1 \over 3}}}

$\therefore$ Density ratio,

{{{\rho _1}} \over {{\rho _2}}} = {{{A_1}/{V_1}} \over {{A_2}/{V_2}}}

= \left( {{{{A_1}} \over {{A_2}}}} \right) \times {\left( {{{{R_2}} \over {{R_1}}}} \right)^3}

= 1:1

Q123

What is the half-life period of a radioactive material if its activity drops to

1 / 16^{\text{th }}

of its initial value in 30 years?

A 9.5 years

B 8.5 years

C 7.5 years

D 10.5 years

Correct Answer

Option C

Solution

$\because$

A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}

\Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{A_0}} \over A} = 16

\Rightarrow {t \over {{T_{1/2}}}} = 4

\Rightarrow {{30} \over {{T_{1/2}}}} = 4

\Rightarrow {T_{1/2}} = {{30} \over 4}

= 7.5

years

Q124

The activity of a radioactive material is

6.4 \times 10^{-4}

curie. Its half life is 5 days. The activity will become

5 \times 10^{-6}

curie after :

A 7 days

B 15 days

C 25 days

D 35 days

Correct Answer

Option D

Solution

$\because$

A = {{{A_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}

\Rightarrow {2^{t/5}} = {{6.4 \times {{10}^{ - 4}}} \over {5 \times {{10}^{ - 6}}}} = 128 = {2^7}

\Rightarrow {t \over 5} = 7

\Rightarrow t = 35

days

Q125

The half life period of a radioactive substance is 60 days. The time taken for

\dfrac{7}{8}

th of its original mass to disintegrate will be :

A 120 days

B 130 days

C 180 days

D 20 days

Correct Answer

Option C

Solution

$\because$

N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}

\Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}}

\Rightarrow {2^{{t \over {{T_{1/2}}}}}} = 8 = {2^3}

\Rightarrow t = 3 \times {T_{1/2}} = 3 \times 60

= 180

days

Q126

A radioactive sample decays

\dfrac{7}{8}

times its original quantity in 15 minutes. The half-life of the sample is

A 5 min

B 7.5 min

C 15 min

D 30 min

Correct Answer

Option A

Solution

N = {{{N_0}} \over {{2^{{t \over {{T_{1/2}}}}}}}}

\Rightarrow {2^{{t \over {{T_{1/2}}}}}} = {{{N_0}} \over N} = {{{N_0}} \over {\left( {{{{N_0}} \over 8}} \right)}} = 8

\Rightarrow {t \over {{T_{1/2}}}} = 3

\Rightarrow {T_{1/2}} = {{15} \over 3} = 5

min

Q127

Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.

A 3 : 4

B 4 : 3

C 1 : 4

D 4 : 1

Correct Answer

Option A

Solution

{E_1} = {E_0}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right) = {E_0} \times {3 \over 4}

{E_2} = {E_0}

$\therefore$

{{{E_1}} \over {{E_2}}} = {3 \over 4}

Q128

Read the following statements : (A) Volume of the nucleus is directly proportional to the mass number. (B) Volume of the nucleus is independent of mass number. (C) Density of the nucleus is directly proportional to the mass number. (D) Density of the nucleus is directly proportional to the cube root of the mass number. (E) Density of the nucleus is independent of the mass number. Choose the correct option from the following options.

A (A) and (D) only.

B (A) and (E) only.

C (B) and (E) only.

D (A) and (C) only.

Correct Answer

Option B

Solution

We know, Radius of nucleus,

r = {r_0}{A^{{1 \over 3}}}

where, A = mass number $\therefore$ Volume

(v) = {4 \over 3}\pi {r^3}

= {4 \over 3}\pi r_0^3\,.\,A

$\therefore$

v \propto A

$\therefore$ Volume is directly proportional to mass number. We know, density

(d) = {m \over v}

= {{z{m_p} + (A - z){m_N}} \over {{4 \over 3}\pi r_0^3{{({A^{1/3}})}^3}}}

= {{z{m_p} + A{m_p} - z{m_p}} \over {{4 \over 3}\pi r_0^3A}}

[ $\because$

{m_p} \simeq {m_N}

]

= {{A{m_p}} \over {{4 \over 3}\pi r_0^3A}}

= {{{m_p}} \over {{4 \over 3}\pi r_0^3}}

$\therefore$ Density of nucleus is independent of mass number.

Q129

Consider the following radioactive decay process

_{84}^{218}A\overset{\alpha}\longrightarrow {A_1}\overset{{{\beta ^ - }}}\longrightarrow {A_2}\overset{\gamma}\longrightarrow {A_3}\overset{\alpha}\longrightarrow {A_4}\overset{{{\beta ^ + }}}\longrightarrow {A_5}\overset{\gamma}\longrightarrow {A_6}

The mass number and the atomic number of A

_6

are given by :

A 210 and 84

B 210 and 80

C 211 and 80

D 210 and 82

Correct Answer

Option B

Solution

${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow}{ }_{82}^{214} A_{4} \stackrel{\beta^{-}}{\longrightarrow}{ }_{83}^{214} A_{2} \stackrel{\gamma}{\longrightarrow}{ }_{83}^{214} A_{3} \stackrel{\alpha}{\longrightarrow}{ }_{81}^{210} A_{4} \stackrel{\beta^{+}}{\longrightarrow}{ }_{80}^{210} A_{5} \stackrel{\gamma}{\longrightarrow}{ }_{80}^{210} A_{6}$ Mass number $=210$ Atomic number $=80$

Q130

An electron of a hydrogen like atom, having

Z=4

, jumps from

4^{\text{th }}

energy state to

2^{\text{nd }}

energy state. The energy released in this process, will be : (Given Rch =

13.6~\mathrm{eV}

) Where R = Rydberg constant c = Speed of light in vacuum h = Planck's constant

A

10.5 ~\mathrm{eV}

B

40.8 ~\mathrm{eV}

C

13.6 ~\mathrm{eV}

D

3.4 ~\mathrm{eV}

Correct Answer

Option B

Solution

The energy difference between the 4th and 2nd energy states of a hydrogen-like atom can be calculated using the formula for the energy levels of a hydrogen-like atom: $\begin{aligned} \Delta \mathrm{E} & =13.6 \mathrm{Z}^2\left(\dfrac{1}{\mathrm{n}_1^2}-\dfrac{1}{\mathrm{n}_2^2}\right) \\\\ \mathrm{Z} & =4 \text{ (hydrogen like atom) } \\ \mathrm{n}_1 & =2, \mathrm{n}_2=4 \\\\ \Delta \mathrm{E} & =13.6(4)^2\left(\dfrac{1}{4}-\dfrac{1}{16}\right) \\\\ & =13.6 \times\left(\dfrac{16-4}{64}\right) \times 16 \\\\ \Delta \mathrm{E} & =13.6 \times \dfrac{12}{64} \times 16 \\\\ \Delta \mathrm{E} & =40.8 \mathrm { eV}\end{aligned}$