Atoms and Nuclei — Page 20 | JEE Physics

Q191

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

{\lambda _n}

,

{\lambda _g}

be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let

{\Lambda _n}

be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)

A

{\Lambda _n} \approx A + {B \over {\lambda _n^2}}

B

{\Lambda _n} \approx A + B{\lambda _n}

C

\Lambda _n^2 \approx A + B\lambda _n^2

D

\Lambda _n^2 \approx \lambda

Correct Answer

Option A

Solution

We know, Wavelength of emitted photon from n2 state to n1 state is

{1 \over \lambda }

= RZ2

\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

Here electron comes from nth state to ground state (n = 1), then the wavelength of photon is ,

{1 \over {{\Lambda _n}}}

= RZ2

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

$\Rightarrow$

\,\,\,

\Lambda

n =

{1 \over {R{Z^2}}}{\left( {1 - {1 \over {{n^2}}}} \right)^{ - 1}}

As n is very large, so using binomial theorem

\Lambda

n =

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

$\Rightarrow$

\,\,\,

\Lambda

n =

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

We know, $\lambda$ n =

{{2\pi r} \over n}

= 2 $\pi$

{\left( {{{{n^2}{h^2}} \over {4{\pi ^2}mZ{C^2}}}} \right)\times{{1 \over n}}}

\therefore\,\,\,

$\lambda$ n $\propto$ n $\Rightarrow$

\,\,\,

n = K $\lambda$ n

\therefore\,\,\,

\Lambda

n =

{1 \over {R{Z^2}}} + {1 \over {R{Z^2}}}\left( {{1 \over {{{\left( {K\,{\lambda _n}} \right)}^2}}}} \right)

Let A =

{1 \over {R{Z^2}}}

and B =

{1 \over {{K^2}R{Z^2}}}

$\Rightarrow$

\,\,\,

\Lambda

n = A +

{B \over {\lambda _n^2}}

Q192

Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively, If initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be :

A 9 : 8

B 1 : 8

C 8 : 1

D 3 : 8

Correct Answer

Option A

Solution

NA = N0

{\left( {{1 \over 2}} \right)^{{t \over {{T_{{1 \over 2}}}}}}}

= N0

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

= N0

{\left( {{1 \over 2}} \right)^6}

NB = N0

{\left( {{1 \over 2}} \right)^{{t \over {{T_{{1 \over 2}}}}}}}

= N0

{\left( {{1 \over 2}} \right)^{{{60} \over {20}}}}

= N0

{\left( {{1 \over 2}} \right)^3}

Decayed nuclei of A = N0 - N0

{\left( {{1 \over 2}} \right)^6}

Decayed nuclei of B = N0 - N0

{\left( {{1 \over 2}} \right)^3}

$\therefore$ Ratio =

{{{N_0} - {N_0}{{\left( {{1 \over 2}} \right)}^6}} \over {{N_0} - {N_0}{{\left( {{1 \over 2}} \right)}^3}}}

=

{{{N_0}\left( {1 - {1 \over {{2^6}}}} \right)} \over {{N_0}\left( {1 - {1 \over {{2^3}}}} \right)}}

=

{{\left( {{{63} \over {{2^6}}}} \right)} \over {\left( {{7 \over {{2^3}}}} \right)}}

=

{9 \over 8}

Q193

Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is :

A n = 1

\to

n = 4

B n = 2

\to

n = 5

C n = 2

\to

n = 4

D n = 2

\to

n = 3

Correct Answer

Option C

Solution

Energy released for tension n = 2 to n = 1 of hydrogen atom

E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

Z = 1, n1 = 1, n2 = 2

E = 13.6 \times 1 \times \left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)

E = 13.6 \times {3 \over 4}eV

= 10.2 eV For He+ ion z = 2 (A) n = 1 to n = 4

E = 13.6 \times {2^2} \times \left( {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{15} \over 4}eV

(B) n = 2 to n = 4

E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{3} \over 4}eV

(C) n = 2 to n = 5

E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right) = 13.6 \times {{21} \over 25}eV

(D) n = 2 to n = 3

E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 13.6 \times {{5} \over 9}eV

So, possible transition is n = 2 $\to$ n = 4

Q194

A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7

\times

1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively :

A 5 days and 10 days

B 10 days and 40 days

C 20 days and 5 days

D 20 days and 10 days

Correct Answer

Option C

Solution

Let number of nuclei present in material A is NA and in meterial B is NB According to question, NA = 2NB We know, activity (A) = $\lambda$ N.

$\therefore$ $\lambda$ ANA = 10 . . . . . . (1) and $\lambda$ B NB = 20 . . . . . . .

(2) $\therefore$ By dividing (1) by (2), we get

{{{\lambda _A}{N_A}} \over {{\lambda _B}{N_B}}}

=

{1 \over 2}

$\Rightarrow$

{{{\lambda _A}} \over {{\lambda _B}}} \times 2

=

{1 \over 2}

[as NA = 2NB] $\Rightarrow$

{{{\lambda _A}} \over {{\lambda _B}}}

=

{1 \over 4}

We know, T

_{{1 \over 2}}

$\propto$

{1 \over \lambda }

$\therefore$

{{{{\left( {{T_{{1 \over 2}}}} \right)}_A}} \over {{{\left( {{T_{{1 \over 2}}}} \right)}_B}}}

=

{{{\lambda _B}} \over {{\lambda _A}}} = 4

$\Rightarrow$

{\left( {{T_{{1 \over 2}}}} \right)_A}

= 4

{\left( {{T_{{1 \over 2}}}} \right)_B}

$\therefore$ By checking options you can see possible value of

{\left( {{T_{{1 \over 2}}}} \right)_A}

= 20 days and

{\left( {{T_{{1 \over 2}}}} \right)_B}

= 5 days

Q195

At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio

{{{R_B}} \over {{R_A}}}

of their activities after time t itself decays with time t as e

-

3t. If the half-life of A is ln2, the half-life of B is :

A 4ln2

B

{{\ln 2} \over 2}

C

{{\ln 2} \over 4}

D 2ln2

Correct Answer

Option C

Solution

We know, Activity (R) = R0 e $-$ $\lambda$ t Given that, at t = 0 RA = RB $\Rightarrow$ R0A e $-$ $\lambda$ Ax0 = R0B e $-$ $\lambda$ Bx0 $\Rightarrow$ R0A = R0B Given that at time t,

{{{R_B}} \over {{R_A}}} = {e^{^{ - 3t}}}

$\Rightarrow$

{{{R_{0B}}\,{e^{ - {\lambda _B}t}}} \over {{R_{0A}}\,{e^{ - {\lambda _A}t}}}}

= e $-$ 3t $\Rightarrow$

{e^{\left( {{\lambda _A} - {\lambda _B}} \right)t}} = {e^{ - 3t}}

$\Rightarrow$ ( $\lambda$ A $-$ $\lambda$ B)t = $-$ 3t $\Rightarrow$ $\lambda$ A $-$ $\lambda$ B = $-$ 3 . . . . .

(1) We know half life,

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

$\therefore$ Half life of A is,

{{t_{{1 \over 2}}} = {{{{\ln }^2}} \over {{\lambda _A}}}}

$\therefore$ $\lambda$ A =

{{\ln 2} \over {{t_{{1 \over 2}}}}}

=

{{\ln 2} \over {\ln 2}}

= 1 From equation (1) we get, $\lambda$ A $-$ $\lambda$ B = $-$ 3 $\Rightarrow$ 1 $-$ $\lambda$ B = $-$ 3 $\Rightarrow$ $\lambda$ B = 4 $\therefore$ Half life of B is

\left( {{t_{{1 \over 2}}}} \right) = {{\ln 2} \over {{\lambda _B}}} = {{\ln 2} \over 4}

Q196

Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -

A 200

B 150

C 400

D 360

Correct Answer

Option A

Solution

at t = 0, A0 =

{{dN} \over {dt}}

= 1600C/s at t = 8s, A = 100 C/s

{A \over {{A_0}}}

=

{1 \over {16}}

in 8 sec Therefor half life is t1/2 = 2 sec $\therefore$ Activity at t = 6 will be 1600

{\left( {{1 \over 2}} \right)^3}

= 200 C/s

Q197

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980

\mathop A\limits^ \circ

. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV

\mathop A\limits^ \circ

)

A 4a0

B 9a0

C 25a0

D 16a0

Correct Answer

Option D

Solution

Energy of photon

= {{12500} \over {980}} = 12.75eV

$\therefore$ Electron will excite to n = 4 Since 'R' $\propto$ n2 $\therefore$ Radius of atom will be 16a0

Q198

\text{ Match the LIST-I with LIST-II }

.tg .tg List - I List - II A.

{ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_0^1 \mathrm{n}

I.

\text{ Chemical reaction }

B.

2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}

II.

\text{ Fusion with +ve } \mathrm{Q} \text{ value }

C.

{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n}

III.

\text{ Fission }

D.

{ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}

IV.

\text{ Fusion with -ve } Q \text{ value }

\text{ Choose the correct answer from the options given below: }

A A-II, B-I, C-III, D-IV

B A-III, B-I, C-II, D-IV

C A-III, B-I, C-IV, D-II

D A-II, B-I, C-IV, D-III

Correct Answer

Option B

Solution

Here’s the correct matching: A.

_0^1\text{n}+_{92}^{235}\text{U}\to_{54}^{140}\text{Xe}+_{38}^{94}\text{Sr}+2\,_0^1\text{n}

–– This is nuclear fission ⇒ III. B.

2\text{H}_2+\text{O}_2\to2\text{H}_2\text{O}

–– This is a chemical reaction ⇒ I. C.

_1^2\text{H}+_1^2\text{H}\to_2^3\text{He}+_0^1\text{n}

–– This D–D fusion releases about 3.3 MeV ⇒ fusion with +ve Q ⇒ II. D.

_1^1\text{H}+_1^3\text{H}\to_1^2\text{H}+_1^2\text{H}

–– The mass of products exceeds reactants ⇒ requires energy ⇒ fusion with –ve Q ⇒ IV.

So the answer is Option B: A–III, B–I, C–II, D–IV.

Q199

Muon (

\mu

-

) is a negatively charged (|q| = |e|) particle with a mass m

\mu

= 200 me, where me is the mass of the electron and e is the electronic charge. If

\mu

-

is bond to a proton to form a hydrogen like atom, identify the correct statements. (A) Radis of the muonic orbit is 200 times smaller than that of the electron. (B) The speed of the

\mu

-

in the nth orbit is

{1 \over {200}}

times that of the electron in the nth orbit. (C) The ionization energy of muonic atom is 200 timesmore than of an hydroen atom. (D) The momentum of the muon in the nth orbit is 200 times more than that of the electron.

A (A), (B), (D)

B (A), (C), (D)

C (B), (D)

D (C), (D)

Correct Answer

Option B

Solution

(i) Radius of the orbit is given by

r = {{{h^2}} \over {4\pi m{e^2}}} \times {{{n^2}} \over Z}

Only, m $\mu$ = 200 me rest are same.

So, statement (A) is correct: The radius of muonic orbit is 200 times smaller than that of the electron. (ii) Velocity of particle in an orbit is given by

v = {{nh} \over {2\pi mr}} = {{nh \times 4\pi m{e^2} \times Z} \over {2\pi m \times {h^2} \times {n^2}}} = {{2{e^2}Z} \over {hn}}

Therefore, speed does not change as it does not depend on mass.

So, statement (B) is incorrect. (iii) Ionisation energy is given by

\Delta {E_n} = - {{2{\pi ^2}m{e^4}{Z^2}} \over {{h^2}{{(4\pi {\varepsilon _0})}^2}}} \times \left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)

Only, m $\mu$ = 200 me rest are same.

So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom. (iv) Since Momentum $\propto$ Energy.

So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.

Q200

Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?

A 8

B 6

C 1

D 4

Correct Answer

Option B

Solution

{E_n} = - 13.6{{{Z^2}} \over {{n^2}}}

Enth of Carbon = E1st of Hydrogen $\Rightarrow$

- 13.6 \times {{{6^2}} \over {{n^2}}} = - 13.6 \times {{{1^2}} \over {{1^2}}}

$\Rightarrow$ n = 6