Atoms and Nuclei — Page 7 | JEE Physics

Q61

An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is :

A 8 : 27

B 4 : 9

C 3 : 2

D 2 : 3

Correct Answer

Option C

Solution

The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

{m_1}{v_1} = {m_2}{v_2} \Rightarrow {{{v_1}} \over {{v_2}}} = {{{m_2}} \over {{m_1}}} \Rightarrow {{{m_2}} \over {{m_1}}} = {8 \over {27}}

Now, since

m = \rho {4 \over 3}\pi {r^3}

Therefore,

{{{m_2}} \over {{m_1}}} = {{\rho {4 \over 3}\pi r_2^3} \over {\rho {4 \over 3}\pi r_1^3}} \Rightarrow {{{m_2}} \over {{m_1}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} \Rightarrow {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} = {8 \over {27}}

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

Thus, ratio of radii of nuclei

{r_1}:{r_2} = 3:2

.

Q62

At some instant, a radioactive sample S1 having an activity 5

\mu

Ci has twice the number of nuclei as another sample S2 which has an activity of 10

\mu

Ci. The half lives of S1 and S2 are :

A 20 years and 5 years, respectively

B 20 years and 10years, respectively

C 5 years and 20 years, respectively

D 10 years and 20 years, respectively

Correct Answer

Option A

Solution

As per question, N1 = 2N2 Also A1 = 5 $\mu$ Ci, A2 = 10 $\mu$ Ci As A = $\lambda$ N =

{{\ln 2} \over {{T_{1/2}}}}N

$\therefore$

{{{A_1}} \over {{A_2}}} = {{{{\left( {{T_{1/2}}} \right)}_2}} \over {{{\left( {{T_{1/2}}} \right)}_1}}} \times {{{N_1}} \over {{N_2}}}

$\Rightarrow$

{{{{\left( {{T_{1/2}}} \right)}_1}} \over {{{\left( {{T_{1/2}}} \right)}_2}}} = {{{N_1}} \over {{N_2}}} \times {{{A_2}} \over {{A_1}}}

= 2 $\times$ 2 = 4 So half life of sample S1 should be four timrs than sample S2.

Q63

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively :

A (0, 1)

B (0.89, 0.28)

C (0.28, 0.89)

D (0, 0)

Correct Answer

Option B

Solution

Applying conservation of momentum : mv + 0 = mv1 + 2mv2 $\Rightarrow$

\,\,\,

v = v1 + 2v2 . . . . . (1) As collision is elastic, So, coefficient of restitution, e = 1

\therefore\,\,\,

e = 1 =

{{velocity\,\,of\,\,separation} \over {Velocity\,\,of\,\,approach}}

$\Rightarrow$

\,\,\,

1 =

{{{v_2} - {v_1}} \over {v - 0}}

$\Rightarrow$

\,\,\,

v = v2 $-$ v1 . . . . .(2) Add (1) and (2), 2v = 3v2 $\Rightarrow$

\,\,\,

v2 =

{{2v} \over 3}

put value of v2 in equation (1), v1 = v $-$ 2v2 = v $-$

{{4v} \over 3}

= $-$

{v \over 3}

\therefore\,\,\,

Fractional loss of energy of neutron. Pd =

{{{k_i} - {k_f}} \over {{k_i}}}

=

{{{1 \over 2}m{v^2} - {1 \over 2}mv_1^2} \over {{1 \over 2}m{v^2}}}

=

{{{v^2} - {{{v^2}} \over 9}} \over {{v^2}}}

=

{8 \over 9}

= 0.89 Applying momentum of conservation, mv + 0 = mv1 + 12mv2 $\Rightarrow$

\,\,\,

v = v1 + 12v2 . . . . . (3) Here also e = 1

\therefore\,\,\,

e = 1 =

{{{v_2} - v{}_1} \over {v - 0}}

$\Rightarrow$

\,\,\,

v = v2 $-$ v1 . . . . . . (4) adding (3) and (4), we get 2v = 13v2 $\Rightarrow$

\,\,\,

v2 =

{{2v} \over {13}}

put this v2 in equation (3), we get v1 = v $-$ 12 $\times$

{{2v} \over {13}}

= $-$

{{11v} \over {13}}

\therefore\,\,\,

Frictional loss pc =

{{{1 \over 2}m{v^2} - {1 \over 2}m{{\left( {{{11} \over {13}}v} \right)}^2}} \over {{1 \over 2}m{v^2}}}

=

{{48} \over {169}}

= 0.28

Q64

The energy required to remove the electron from a singly ionized Helium atom is

2.2

times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is :

A

20

eV

B

34

eV

C

79

eV

D

109

eV

Correct Answer

Option C

Solution

Energy required to remove e $-$ from singly ionized Helium atom E1 =

{{13.6{z^2}} \over {{n^2}}}

=

{{13.6 \times {2^2}} \over {{1^2}}}

= 54.4 eV. Let, E2 = energy required to remove e $-$ from He $-$ atom

\therefore\,\,\,\,

According to q vertion. E1 = 2.2 E2

\therefore\,\,\,\,

E2 =

{{54.4} \over {2.2}}

= 24.72 eV

\therefore\,\,\,\,

Total energy required to ionize Helium atom completely = (54.4 + 24.72) eV = 79.12 eV

Q65

Two radioactive materials A and B have decay constants 10

\lambda

and

\lambda

, respectively. It initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be 1/e after a time :

A 1/9

\lambda

B 11/10

\lambda

C 1/10

\lambda

D 1/11

\lambda

Correct Answer

Option A

Solution

N1 = N0e–10 $\lambda$ t ; N2 = N0e– $\lambda$ t

{1 \over e} = {{{N_1}} \over {{N_2}}} = {e^{ - 9\lambda t}}

\Rightarrow 9\lambda t = 1

\Rightarrow t = {1 \over {9\lambda }}

Q66

In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength

\lambda

. When the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of

\lambda

? (Given : H = 6.63 × 10–34 Js; c = 3 × 108 ms –1)

A 10.8 nm

B 12.3 nm

C 9.4 nm

D 11.4 nm

Correct Answer

Option A

Solution

{{hc} \over \lambda } = 13.6\,ev(g)\left\{ {1 - {1 \over {16}}} \right\}

{{1240\,eV} \over \lambda } = {{15} \over {16}} \times 9 \times 13.6\,eV

\lambda = {{1240 \times 16} \over {15 \times 9 \times 13.6}} = 10.8\,nm

Q67

An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength

\lambda

, energy

E = {{1240\,eV} \over {\lambda (in\,nm)}}

) :

A n = 4

B n = 7

C n = 5

D n = 6

Correct Answer

Option C

Solution

\Delta {E_n} = - {{{E_0}{Z^2}} \over {{n^2}}}

Let it start from n to m and from m to ground. Then

13.6 \times 4\left| {1 - {1 \over {{m^2}}}} \right| = {{hc} \over {30.4\,nm}}

\Rightarrow 1 - {1 \over {{m^2}}} = 0.7498 \Rightarrow 0.25 = {1 \over {{m^2}}}

$\therefore$ m = 2, and now

13.6 \times 4\left( {{1 \over 4} - {1 \over {{n^2}}}} \right) = {{hc} \over {108.5 \times {{10}^{ - 9}}}}

n = 5

Q68

Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65

\mathop A\limits^o

). The de-Broglie wavelength of this electron is :

A 6.6

\mathop A\limits^o

B 3.5

\mathop A\limits^o

C 9.7

\mathop A\limits^o

D 12.9

\mathop A\limits^o

Correct Answer

Option C

Solution

For second excited state n = 3

mvr = {{3h} \over {2\pi }}

........(1)

mv = {h \over \lambda }

.........(2) Dividing (1) by (2), we get

r = {{3\lambda } \over {2\pi }}

$\Rightarrow$

\lambda = {{2\pi r} \over 3}

=

{{2 \times 3.14 \times 4.65} \over 3}

= 9.7

\mathop A\limits^o

Q69

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths,

{{{\lambda _1}} \over {{\lambda _2}}}

, of the photons emitted in this process is :

A

{{22} \over 5}

B

{7 \over 5}

C

{9 \over 7}

D

{{20} \over 7}

Correct Answer

Option D

Solution

n = 1 (Ground state) n = 2 (First excitate state) n = 3 (Second excitate state) n = 4 (Third excitate state)

{{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} \right)

{{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)

{{{\lambda _2}} \over {{\lambda _1}}} = {{\left( {{7 \over {9 \times 16}}} \right)} \over {\left( {{5 \over {9 \times 4}}} \right)}}

=

{7 \over {20}}

$\therefore$

{{{\lambda _1}} \over {{\lambda _2}}} = {{20} \over 7}

Q70

The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be

A 2.25R

B

3 \mathrm{R}

C

\dfrac{\mathrm{R}}{3}

D

9 \mathrm{R}

Correct Answer

Option A

Solution

$r \propto \dfrac{n^{2}}{Z}$

\begin{aligned} & \frac{r_{2 { nd }}}{r_{3 \mathrm{rd}}}=\left(\frac{n_{2}}{n_{3}}\right)^{2} \\\\ & \Rightarrow \frac{R}{r_{3 r d}}=\left(\frac{2}{3}\right)^{2} \\\\ & \Rightarrow r_{3 \mathrm{rd}}=\frac{9}{4} R \\\\ & =2.25 R \end{aligned}