Atoms and Nuclei — Page 10 | JEE Physics

Q91

The half-life of Au198 is 2.7 days. The activity of 1.50 mg of Au198 if its atomic weight is 198 g mol

-

1 is, (Na = 6

\times

1023/mol).

A 240 Ci

B 357 Ci

C 252 Ci

D 535 Ci

Correct Answer

Option B

Solution

Activity,

A = \lambda N

Where,

N = n{N_A}

{N_A} = 6 \times {10^{23}}

/mol

= {{6 \times {{10}^{23}}} \over {3.7 \times {{10}^{10}}}}

Ci Here,

{A_0} = \lambda {N_0}

We know,

{T_{1/2}} = {{\ln (2)} \over \lambda }

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

= {{\ln (2)} \over {2.7 \times 3600 \times 24}}

$\therefore$

{A_0} = {{\ln (2)} \over {2.7 \times 3600 \times 24}} \times

n \times {N_A}

= {{\ln (2)} \over {2.7 \times 3600 \times 24}} \times {{1.5 \times {{10}^{ - 3}}} \over {198}} \times {{6 \times {{10}^{23}}} \over {3.7 \times {{10}^{10}}}}

= 366

Ci $\therefore$ Nearest answer is 357 Ci.

Q92

Imagine that the electron in a hydrogen atom is replaced by a muon (

\mu

). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :

A 13.6 eV

B 2815.2 eV

C 331.2 eV

D 27.2 eV

Correct Answer

Option B

Solution

mm = 207 me In hydrogen atom one electron present.

Now that the electron in hydrogen atom is replaced by a muon ( $\mu$ ).

We know, Energy(E) =

- {{{e^4}m} \over {8\varepsilon _0^2{n^2}{h^2}}}

For electron, Ee =

- {{{e^4}{m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}

= 13.6 eV For muon, Em =

- {{{e^4}\left( {207} \right){m_e}} \over {8\varepsilon _0^2{n^2}{h^2}}}

= 13.6 $\times$ 207 = 2815.2 eV

Q93

A radioactive sample disintegrates via two independent decay processes having half lives

T_{1/2}^{(1)}

and

T_{1/2}^{(2)}

respectively. The effective half-life T1/2 of the nuclei is :

A None of the above

B

{T_{1/2}} = T_{1/2}^{(1)} + T_{1/2}^{(2)}

C

{T_{1/2}} = {{T_{1/2}^{(1)}T_{1/2}^{(2)}} \over {T_{1/2}^{(1)} + T_{1/2}^{(2)}}}

D

{T_{1/2}} = {{T_{1/2}^{(1)} + T_{1/2}^{(2)}} \over {T_{1/2}^{(1)} - T_{1/2}^{(2)}}}

Correct Answer

Option C

Solution

{\left( {{{dN} \over {dt}}} \right)_1} = N{\lambda _1},{\left( {{{dN} \over {dt}}} \right)_2} = N{\lambda _2}

{{dN} \over {dt}} = {\left( {{{dN} \over {dt}}} \right)_1} + {\left( {{{dN} \over {dt}}} \right)_2}

N{\lambda _{eff}} = N{\lambda _1} \times N{\lambda _2}

{1 \over {{T_{eff}}}} = {1 \over {{T_1}}} + {1 \over {{T_2}}}

{T_{eff}} = {{{T_1}{T_2}} \over {{T_1} + {T_2}}}

Q94

The decay of a proton to neutron is :

A always possible as it is associated only with

\beta

+ decay

B possible only inside the nucleus

C not possible as proton mass is less than the neutron mass

D not possible but neutron to proton conversation is possible

Correct Answer

Option B

Solution

Positron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in which a proton inside a nucleus is converted into a neutron while releasing a positron and an electron neutrino.

So, decay of a proton to neutron is possible only inside the nucleus.

Free proton cannot decay to neutron as mass of proton is less compared to neutron so to decay into higher mass, proton need extra energy which free proton can’t get, only proton inside nucleus can get.

Q95

A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains ? (Take ln 3 = 1.1)

A 740 years

B 1110 years

C 700 years

D 340 years

Correct Answer

Option A

Solution

The given situation can be shown as Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, $\lambda$ a and $\lambda$ b.

It means that $\because$

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

where, t1/2(a) = 700 yr and t1/2(b) = 1400 yr

{\lambda _a} = {{\ln 2} \over {700}}y{r^{ - 1}}

and

{\lambda _b} = {{\ln 2} \over {1400}}y{r^{ - 1}}

$\because$

{\lambda _{total}} = {\lambda _a} + {\lambda _b}

= \left( {{{\ln 2} \over {700}} + {{\ln 2} \over {1400}}} \right)y{r^{ - 1}} = \ln 2\left( {{1 \over {700}} + {1 \over {1400}}} \right)y{r^{ - 1}}

= \left( {{{3\ln 2} \over {1400}}} \right)y{r^{ - 1}}

Suppose the initial number of radioactive nuclei was N0. $\therefore$

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

where, N = number of nuclei present at time = t and N0 = number of nuclei present at time = 0

\Rightarrow {{{N_0}} \over 3} = {N_0}{e^{ - \lambda t}}

or

{{{N_0}} \over 3} = {N_0}{N_0}{e^{ - {\lambda _{total}}t}} \Rightarrow {1 \over 3} = {e^{ - {\lambda _{total}}t}}

Taking log on both the sides of above equation, we get

\ln \left( {{1 \over 3}} \right) = \ln ({e^{ - {\lambda _{total}}t}})

\Rightarrow \ln \left( {{1 \over 3}} \right) = - {\lambda _{total}}t

\Rightarrow 1.1 = {{3 \times 0.693} \over {1400}} \times t

$\Rightarrow$ t $\approx$ 740 yr

Q96

A nucleus of mass M emits

\gamma

-ray photon of frequency 'v'. The loss of internal energy by the nucleus is : [Take 'c' as the speed of electromagnetic wave]

A hv

B

hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]

C

hv\left[ {1 - {{hv} \over {2M{c^2}}}} \right]

D 0

Correct Answer

Option B

Solution

Energy of $\gamma$ -ray, E $\gamma$ = hv and momentum of $\gamma$ -ray,

{p_\gamma } = {h \over \lambda }

.... (i) As,

{p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}

..... (ii) From Eqs. (i) and (ii), we get

{p_\gamma } = {{hv} \over c} = {h \over \lambda }

[ $\because$

\lambda = {c \over v}

] Since, during the emission of $\gamma$ -ray photon, momentum is conserved. $\therefore$ p $\gamma$ + pdecayed nuclei = 0 $\Rightarrow$ p $\gamma$ = pdecayed nuclei

\Rightarrow {{hv} \over c}

= pdecayed nuclei ..... (iii) Kinetic energy of decayed nuclei,

KE = {1 \over 2}M{v^2} = {{(p_{decayed\,nuclei}^2)} \over {2M}}

..... (iv) From Eqs. (iii) and (iv), we get

\Rightarrow KE = {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2}

$\therefore$ Loss in internal energy = E $\gamma$ + KEdecayed nuclei

= hv + {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2} = hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]

Q97

A nucleus with mass number 184 initially at rest emits an

\alpha

-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the

\alpha

-particle.

A 5.5 MeV

B 5.0 MeV

C 5.38 MeV

D 0.12 MeV

Correct Answer

Option C

Solution

{k_\alpha } + {k_N} = 5.5

k = {{{p^2}} \over {2m}}

\Rightarrow {{{k_\alpha }} \over {{k_N}}} = {{180} \over 4} = 45

\Rightarrow {k_\alpha } = {{45} \over {46}} \times 5.5

MeV = 5.38 MeV

Q98

Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as : (where

\lambda

is the decay constant)

A

{1 \over 2}{{\ln 2} \over \lambda }

B

{{\ln 2} \over \lambda }

C

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

D

{{\ln {3 \over 2}} \over \lambda }

Correct Answer

Option D

Solution

{{3{N_0}} \over 4} = {N_0}{e^{ - \lambda {t_1}}}

{{{N_0}} \over 2} = {N_0}{e^{ - \lambda {t_2}}}

\ln (3/4) = - \lambda {t_1}

..... (i)

\ln (1/2) = - \lambda {t_2}

..... (i)

\ln (3/4) - \ln (1/2) = \lambda ({t_2} - {t_1})

....(i)

\Delta t = {{\ln (3 /2)} \over \lambda }

Q99

The half-life of

{}^{198}Au

is 3 days. If atomic weight of

{}^{198}Au

is 198 g/mol then the activity of 2 mg of

{}^{198}Au

is [in disintegration/second] :

A 2.67

\times

1012

B 6.06

\times

1018

C 32.36

\times

1012

D 16.18

\times

1012

Correct Answer

Option D

Solution

A = $\lambda$ N

\lambda = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over {3 \times 24 \times 60 \times 60}}

sec $-$ 1 = 2.67 $\times$ 10 $-$ 6 sec $-$ 1 N = Number of atoms in 2 mg Au

= {{2 \times {{10}^{ - 3}}} \over {198}} \times 6 \times {10^{23}}

= 6.06 $\times$ 1015

A = \lambda N = 1.618 \times {10^{13}} = 16.18 \times {10^{12}}

dps

Q100

If 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N0) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as : [

\lambda

is the radioactive decay constant]

A

-

\lambda

(1

-

e

-

\lambda

t)

B

\lambda

(1

-

e

-

\lambda

t)

C

\lambda

e

-

\lambda

t

D

-

\lambda

e

-

\lambda

t

Correct Answer

Option C

Solution

N = N0e $-$ $\lambda$ t Nd = N0 $-$ N Nd = N0 (1 $-$ e $-$ $\lambda$ t)

{{{N_d}} \over {{N_0}}} = f = 1 - {e^{ - \lambda t}}

$\Rightarrow$

{{df} \over {dt}} = \lambda {e^{ - \lambda t}}