Physics Ncert Solutions Class 12th

13.29 It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_1$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 – 0 = 1.088 MeV = 1.088 $*{10}^6$  eV = 1.088 $*{10}^6*1.6*{10}^{-19}$ J

= 1.7408 $*{10}^{-13}$ J

We know, $E_1=h{\nu }_1$ , where

${\nu }_1$ = Frequency of radiation radiated by ${\gamma }_1$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $*{10}^{-34}$ Js

Hence,  ${\nu }_1$ = $\frac{E_1}{h}$  = $\frac{1.7408*{10}^{-13}}{6.6*{10}^{-34}}$  = 2.637 $*{10}^{20}$ Hz

It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_2$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to ${\gamma }_2$  decay is given as:

$E_2$ = 0.412 – 0 = 0.412 MeV = 0.412 $*{10}^6$  eV = 0.412 $*{10}^6*1.6*{10}^{-19}$  J

= 6.592 $*{10}^{-14}$ J

We know, $E_2=h{\nu }_2$ , where

${\nu }_2$ = Frequency of radiation radiated by ${\gamma }_2$ decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $*{10}^{-34}$ Js

Hence, ${\nu }_2$  = $\frac{E_2}{h}$ = $\frac{6.592*{10}^{-14}}{6.6*{10}^{-34}}$  = 9.987 $*{10}^{19}$  Hz

It can be observed from the given $\gamma -$ decay diagram that ${\gamma }_3$ decays from 1.008 MeV energy level to the 0.412 MeV energy level.

Hence the energy corresponding to  decay is given as:

 = 1.088 - 0.412 = 0.676 MeV = 0.676 $*{10}^6$  eV = 0.676 $*{10}^6*1.6*{10}^{-19}$  J

= 1.0816 $*{10}^{-13}$  J

We know, $E_3=h{\nu }_3$ , where

${\nu }_3$ = Frequency of radiation radiated by ${\gamma }_3$  decay

$h=\mathrm{P}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{c}{\mathrm{k}}^{\mathrm{\text{'}}}\mathrm{s}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}$ = 6.6 $*{10}^{-34}$  Js

 Hence, ${\nu }_3$  = $\frac{E_3}{h}$  = $\frac{1.0816*{10}^{-13}}{6.6*{10}^{-34}}$ = 1.639 $*{10}^{20}$ Hz

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) = 197.968233 u

Mass of ( ${}_{79}{}^{198}\mathrm{A}\mathrm{u}$ ) =197.966760 u

Energy of the highest level is given as:

E = $\left[\mathrm{m}\left({}_{79}{}^{198}\mathrm{A}\mathrm{u}\right)-\mathrm{m}\left({}_{80}{}^{198}Hg\right)\mathrm{}\right]c^2$

= $\left[197.968233-197.966760\right]c^2$ u

= 1.473 $*{10}^{-3}{c}^2$ u

= 1.473 $*{10}^{-3}*931.5$  MeV

= 1.3720995 MeV

${\beta }_1^{-}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{s}$  from 1.3720995 MeV level to 1.088 MeV level

Hence maximum kinetic energy of the ${\beta }_1^{-}$ particles = 1.3720995 - 1.088 = 0.2840995 MeV

${\beta }_2^{-}$ from 1.3720995 MeV level to 0.412 MeV level

Hence maximum kinetic energy of the ${\beta }_2^{-}$ particles = 1.3720995 – 0.412 = 0.9600995 MeV

13.25 Half life of ${}_{15}{}^{32}P{,T}_{1/2}$ = 14.3 days

Half life of ${}_{15}{}^{33}P,T_{1/2}$  = 25.3days and ${}_{15}{}^{33}P$  decay is 10% of the total amount of decay

The source has initially 10 % of ${}_{15}{}^{33}P$ nucleus and 90% of ${}_{15}{}^{32}P$  nucleus

Suppose after t days, the source has 10% of ${}_{15}{}^{32}P$  nucleus and 90% of ${}_{15}{}^{33}P$  nucleus

Initially:

Number of ${}_{15}{}^{33}P$ nucleus = N

Number of ${}_{15}{}^{32}P$  nucleus = 9N

Finally:

Number of ${}_{15}{}^{33}P$ nucleus = 9N'

Number of ${}_{15}{}^{32}P$  nucleus = N'

For ${}_{15}{}^{32}P$ nucleus, the number ratio, $\frac{N\text{'}}{9N}$  = ${\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

N' = 9N ${\left(2\right)}^{-t/14.3}$ …………….(1)

For ${}_{15}{}^{33}P$  nucleus, the number ratio, $\frac{9N\text{'}}{N}$ = ${\left(\frac{1}{2}\right)}^{t/T_{1/2}}$

9N' = N ${\left(2\right)}^{-t/25.3}$  …………….(2)

On dividing equation (1) by equation (2), we get:

$\frac{1}{9}$ = 9 $*2^{(\frac{t}{25.3}-\frac{t}{14.3})}$

$\frac{1}{81}$  = $2^{\left(\frac{-11t}{25.3*14.3}\right)}$

Taking log on both sides

log 1 – log 81 = $\frac{-11t}{25.3*14.3}$ log2

0 – 1.908 = ( $\frac{-11t}{361.79}$ ) $*0.301$

t = 208.5 days

Hence, it will take about 208.5 days for 90% decay of $P_{15}^{33}$

13.22 Let the amount on energy released during the electron capture process be $Q_1$ . The nuclear reaction can be written as:

 $e^{+}$+ ${}_Z{}^{A}X$ $\to {}_{z-1}{}^{A}Y+v+Q_1$ …………………….(1)

Let the amount of energy released during the positron capture process be $Q_2$. The nuclear reaction can be written as:

 ${}_Z{}^{A}X$ $\to {}_{z-1}{}^{A}Y+e^{+}+v$ +$Q_2$ …………………….(2)

Let us assume

$m_N\left({}_Z{}^{A}X\right)$ = Nuclear mass of ${}_Z{}^{A}X$

$m_N\left({}_Z{}^{A}X\right)$ = Atomic mass of ${}_Z{}^{A}X$

$m_N\left({}_{Z-1}{}^{A}Y\right)$ = Nuclear mass of ${}_{Z-1}{}^{A}Y$

$m_N\left({}_{Z-1}{}^{A}Y\right)$ = Atomic mass of ${}_{Z-1}{}^{A}Y$

$m_e$ = mass of the electron

c = speed of light

Q-value of the electron capture reaction is given as:

$Q_1$ = [ 

= [ $m\left({}_Z{}^{A}X\right)-Z{m}_e+m_e-m\left({}_{Z-1}{}^{A}Y\right)+(Z-1)m_e]c^2$

= [$m\left({}_Z{}^{A}X\right)-m\left({}_{Z-1}{}^{A}Y\right)]c^2$  ………………….(3)

Q-value of the positron capture reaction is given as:

$Q_2$ = [ $m_N\left({}_Z{}^{A}X\right)-m_N\left({}_{Z-1}{}^{A}Y\right)-m_e]c^2$