Nuclei

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 $\times {10}^6$  eV = 1.088 $\times {10}^6\times 1.6\times {10}^{-19}$ J

= 1.7408 $\times {10}^{-13}$ J

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

${\nu }_1$ = Frequency of radiat

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 ${}_{15}{}^{33}P$ nucleu

13.23 Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotope ${}_{12}{}^{24}\mathit{M}\mathit{g}$ ,= 23.98504u

Mass of magnesium isotope  ${}_{12}{}^{25}Mg$ ,$m_2$= 24.98584u

Mass of magnesium isotope  ${}_{12}{}^{26}Mg$ ,$m_3$= 25.98259u

Let the abundance of magnesium isotope  ${}_{12}{}^{24}Mg$ be ${\eta }_1$ = 78.99 %

Let the abundance of magnesium isotope  ${}_{12}{}^{25}Mg$ be ${\eta }_2$= x %

T

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 ass

13.21 We have the expression for nuclear radius as:

R = $R_0{A}^{1/3}$

Where $R_0$ = constant

A = mass number of nucleus

Let m be the average mass of the nucleus, hence mass of the nucleus = mA

Nuclear matter density $\rho$ can be written as

 $\rho =\frac{Massofthenucleus}{Volumeofthenucleus}$ = $\frac{mA}{\frac{4}{3}\pi R^3}$ = $\frac{3mA}{4\pi {\left(R_0{A}^{\frac{1}{3}}\right)}^3}$ = $\frac{3mA}{4\pi R_0^{3}A}$ = $\frac{3m}{4\pi R_0^3}$

Hence, the nuclear mass d

13.20 When two deuterons collide head-on, the distance between their centers, d is given as

d = Radius of 1st deuteron + Radius of 2nd deuteron

Radius of the deuteron nucleus = 2 fm = 2 $\times {10}^{-15}$ m

Hence d = 2 $\times {10}^{-15}+2\mathrm{}\times {10}^{-15}$ = 4 $\times {10}^{-15}$ m

Charge on a deuteron nucleus = Charge on an electron = 1.6 $\times {10}^{-19}$ C

Potential ener

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 = 
$\frac{6.023\times {10}^{23}}{2}$ $\times 2\times {10}^3$atoms = 6.023 $\times {10}^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27

13.18 Half life of the fuel in the fission reactor, 
= 5 years = 5$\times 365\times 24\times 60\times 60s$

= 157.68 $\times {10}^{6}s$

We know that in the fission of 1 g of ${}_{92}{}^{235}\mathit{U}$
 , the energy released = 200 MeV

1 mole i.e. 235 gm of ${}_{92}{}^{235}U$ contains 6.023 $\times {10}^{23}$atoms

Therefore 1 gm of ${}_{92}{}^{235}\mathit{U}$
 contains =$\frac{6.023\times {10}^{23}}{235}$= 2.563 $\times {10}^{21}$ atoms

The total energy Q generated per gm of &n