Nuclei

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:

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

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

= [ $m\left({}_Z{}^{A}X\right)-m\left({}_{Z-1}{}^{A}Y\right)-2m_e]c^2$ ……………(4)

It can be inferred that if$Q_2>0$ , then $Q_1>0$ , also if  means that $Q_2>0$.

In other words, this means that if ${\beta }^{+}$ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because Q-value must be positive for an energetically-allowed nuclear reaction.

13.19 The given fusion reaction is

$\mathit{}$ ${}_1{}^{2}H$+ ${}_1{}^{2}H$ $\to$ ${}_2{}^3\mathit{H}\mathit{e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023
 atoms

Hence 2 kg of deuterium contains = 
$\frac{6.023*{10}^{23}}{2}$ $*2*{10}^3$atoms = 6.023 $*{10}^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E = 
$\frac{3.27}{2}$ $*$6.023$*{10}^{26}$  MeV

=  $\frac{3.27}{2}$$*$ 6.023 $*{10}^{26}*{10}^6*1.6*{10}^{-19}$ J = 1.5756$*{10}^{14}$ J

Power of the electric bulb, P = 100 W = 100 J/s

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756*{10}^{14}}{100}$ seconds = 1.5756 $*{10}^{12}$secs

= 49.96 $*{10}^3$ years