Atoms and Nuclei — Page 9 | JEE Physics

Q81

The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

A 35.8 nm

B 11.4 nm

C 8.6 nm

D 24.2 nm

Correct Answer

Option B

Solution

So, ionisation energy = (13.6 Z2) eV = 9 $\times$ 13.6 eV

{1 \over \lambda } = R{Z^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right)

= 1.09 $\times$ 107 $\times$ 9 $\times$

{8 \over 9}

= 11.4 nm

Q82

In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = 6.023

\times

1026 per kilo mole and 1 eV = 1.6 × 10–19 J. The power output of the reactor is close to

A 125 MW

B 60 MW

C 54 MW

D 35 MW

Correct Answer

Option B

Solution

Number of uranium atoms in 2 kg =

{{2 \times 6.023 \times {{10}^{26}}} \over {235}}

Energy from one atom is 200 × 106 e.V. hence total energy from 2 kg uranium =

{{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}

eV =

{{2 \times 6.023 \times {{10}^{26}}} \over {235}} \times 200 \times {10^6}

$\times$ 1.6 × 10–19 J 2 kg uranium is used in 30 days hence energy received per second or power is Power =

{{2 \times 6.023 \times {{10}^{26}} \times 200 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}} \over {235 \times 30 \times 24 \times 3600}}

= 63.2 × 106 watt or 63.2 Mega Watt

\simeq

60 MW

Q83

In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to :

A

{1 \over n}

B

{1 \over {{n^2}}}

C

{1 \over {{n^3}}}

D

{1 \over {{n^4}}}

Correct Answer

Option C

Solution

In hydrogen atom, En =

{{ - {E_0}} \over {{n^2}}}

Where E0 is Ionisation Energy of H.

For transition from (n + 1) to n, the energy of emitted radiation is equal to the difference in energies of levels.

\Delta

E = En+1 - En =

{E_0}\left( {{1 \over {{n^2}}} + {1 \over {{{\left( {n + 1} \right)}^2}}}} \right)

=

{E_0}\left( {{{{{\left( {n + 1} \right)}^2} - {n^2}} \over {{n^2}{{\left( {n + 1} \right)}^2}}}} \right)

= E0

\left( {{{2n + 1} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)

= E0

\left( {{{n\left( {2 + {1 \over n}} \right)} \over {{n^4}{{\left( {1 + {1 \over n}} \right)}^2}}}} \right)

Since n >>> 1 Hence,

{{1 \over n} \simeq 0}

=

{E_0}\left( {{2 \over {{n^3}}}} \right)

As

\Delta

E = h $\nu$ $\therefore$ h $\nu$ =

{E_0}\left( {{2 \over {{n^3}}}} \right)

$\Rightarrow$ $\nu$ $\propto$

{1 \over {{n^3}}}

Q84

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by :

A

\vec{\mu}_{\mathrm{L}}=\dfrac{\overrightarrow{\mathrm{eL}}}{2 \mathrm{~m}}

B

\vec{\mu}_{\mathrm{L}}=-\dfrac{\overrightarrow{\mathrm{eL}}}{2 \mathrm{~m}}

C

\vec{\mu}_{l}=-\dfrac{\overrightarrow{e L}}{\mathrm{~m}}

D

\vec{\mu}_{l}=\dfrac{2 \overrightarrow{\mathrm{eL}}}{\mathrm{m}}

Correct Answer

Option B

Solution

$\because$

\overrightarrow \mu = {{q\overrightarrow L } \over {2m}}

\Rightarrow \overrightarrow \mu = {{ - e\overrightarrow L } \over {2m}}

Q85

The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3

\times

10–15)A1/3 m. It follows that the mass density of a nucleus is of the order of : (Mprot.

\cong

Mneut

\simeq

1.67

\times

10–27 kg)

A 1024 kg m–3

B 1010 kg m–3

C 1017 kg m–3

D 103 kg m–3

Correct Answer

Option C

Solution

R = (1.3 \times {10^{ - 15}}){A^{{1 \over 3}}}

We know,

m = pV

$\Rightarrow$

p = {m \over V}

$\Rightarrow$

p = {{{m_p}A} \over {{4 \over 3}\pi {R^3}}}

p = {{{m_p}A} \over {{4 \over 3}\pi \times {{(1.3 \times {{10}^{ - 15}})}^3}A}}

p \approx {10^{17}}kg/m

Q86

According to Bohr atomic model, in which of the following transitions will the frequency be maximum?

A n = 2 to n = 1

B n = 3 to n = 2

C n = 4 to n = 3

D n = 5 to n = 4

Correct Answer

Option A

Solution

Let, nf, ni be the final and initial orbit. As we know that,

{1 \over \lambda } = 1.09 \times {10^7}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]

Now, checking for each option, we get (a)

{1 \over \lambda } \propto \left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right] = \left[ {{1 \over 9} - {1 \over {16}}} \right] = 0.05

.... (i) (b)

{1 \over \lambda } \propto \left[ {{1 \over 1} - {1 \over 4}} \right] = 0.75

.... (ii) (c)

{1 \over \lambda } \propto \left[ {{1 \over {16}} - {1 \over {25}}} \right] = 0.0225

.... (iii) (d)

{1 \over \lambda } \propto \left[ {{1 \over 4} - {1 \over 9}} \right] = 0.14

.... (iv) The option (b) has highest value. Since, frequency,

f = {c \over \lambda } \Rightarrow f \propto {1 \over \lambda }

$\therefore$ Frequency will be maximum for transition n = 2 to n = 1.

Q87

Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio

{{{N_1}} \over {{N_2}}}

will be equal to :

A

{1 \over 8}

B

{3 \over 1}

C

{1 \over 3}

D

{8 \over 1}

Correct Answer

Option D

Solution

Let Half life of x = t then half life of y = 2t when 3 half life of y is completed then 6 half life of x is completed.

$\therefore$ Now x have =

{{{N_1}} \over {{2^6}}}

nuclei and y have =

{{{N_2}} \over {{2^3}}}

nuclei From question,

{{{N_1}} \over {{2^6}}} = {{{N_2}} \over {{2^3}}}

\Rightarrow {{{N_1}} \over {{N_2}}} = 8

Q88

If

\lambda

1 and

\lambda

2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of

\lambda

1 :

\lambda

2 is :

A 7 : 135

B 7 : 108

C 1 : 9

D 1 : 3

Correct Answer

Option A

Solution

For Lyman series n1 = 1, n2 = 4

{1 \over {{\lambda _1}}} = R\left[ {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right]

For paschen series n1 = 3, n2 = 4

{1 \over {{\lambda _2}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]

$\therefore$

{{{\lambda _1}} \over {{\lambda _2}}} = {{\left[ {{1 \over 9} - {1 \over {16}}} \right]} \over {\left[ {1 - {1 \over {16}}} \right]}} = {7 \over {9 \times 15}}

$\Rightarrow$

{{{\lambda _1}} \over {{\lambda _2}}} = {7 \over {135}}

Q89

A radioactive sample is undergoing

\alpha

decay. At any time t1, its activity is A and another time t2, the activity is

{A \over 5}

. What is the average life time for the sample?

A

{{\ln 5} \over {{t_2} - {t_1}}}

B

{{\ln ({t_2} + {t_1})} \over 2}

C

{{{t_1} - {t_2}} \over {\ln 5}}

D

{{{t_2} - {t_1}} \over {\ln 5}}

Correct Answer

Option D

Solution

Let initial activity be A0 A = A0 e $-$ $\lambda$ t1 ........(i)

{A \over 5}

= A0 e $-$ $\lambda$ t2 .......(ii) (i)

\div

(ii) 5 = e $\lambda$ (t2 $-$ t1) $\lambda$ =

{{\ln 5} \over {{t_2} - {t_1}}} = {1 \over \tau }

\tau = {{{t_2} - {t_1}} \over {\ln 5}}

Q90

Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 minutes.

A 40 minutes

B 60 minutes

C 13 minutes

D 20 minutes

Correct Answer

Option D

Solution

{T_{1/2}} = 20 \Rightarrow {{\ln 2} \over \lambda } = 20

min

\Rightarrow \lambda = {{\ln 2} \over {20(\min )}}

$\because$

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

{{{N_t}} \over {{N_0}}} = {e^{ - \lambda {t_1}}} \Rightarrow 0.67 = {e^{ - \lambda {t_1}}}

\Rightarrow \ln (0.67) = - \lambda {t_1}

\Rightarrow \ln \left( {{{100} \over {67}}} \right) = \lambda {t_1}

\Rightarrow {t_1} = {{\ln \left( {{{100} \over {67}}} \right) \times 20(\min )} \over {(\ln 2)}}

Similarly,

{t_2} = {{\ln \left( {{{100} \over {34}}} \right) \times 20(\min )} \over {(\ln 2)}}

{t_2} - {t_1} = 19.57

min

\approx 20

min.