Class 12th

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 density is independent of A. It is nearly constant

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 $*{10}^{-15}$ m

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

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

Potential energy of two-deuteron system: $<math><mi>V</mi></math>$ = $\frac{e^2}{4\pi {?}_{0}d}$$$

Where ${?}_0$= permittivity of free space

 = 9 $*{10}^9$ N $m^2{C}^{-1}$

V = $\frac{{(1.6*{10}^{-19})}^2*9\mathrm{}*{10}^9}{4\mathrm{}*{10}^{-15}}$ J =$\frac{{(1.6*{10}^{-19})}^2*9\mathrm{}*{10}^9}{4\mathrm{}*{10}^{-15}*1.6*{10}^{-19}}$ eV = 360 $*{10}^3$
 eV = 360 keV

Hence the height of the potential barrier of the two-deuteron system is 360 keV.

1.11

(a) When polythene rubbed against wool, a number of electrons get transferred from wool to polythene. Hence, wool becomes positively charged and polythene negatively charged.

Amount of charge on the polythene piece, q = -3 $*{10}^{-7}$ C.

Amount of charge of 1 electron e = -1.6 $*10-19$

So number of electron transferred from wool to polythene

=
-$\frac{3\mathrm{}*{10}^{-7}}{-1.6\mathrm{}*{10}^{-19}}$1.875$*{10}^{12}$

(b) Since electron has a mass, so there will be transfer of mass also.

Mass of single electron,  $<math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub></math>$ = 9.1 $*{10}^{-31}$ kg

Total mass transferred from wool to polythene = 1.875 $*{10}^{12}*$9.1 $*{10}^{-31}$ kg

= 1.706 $*{10}^{-18}$kg$?$ negligible

Hence a negligible amount of mass is transferred from wool to polythene.

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

= 157.68 $*{10}^{6}s$

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 $*{10}^{23}$atoms

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

The total energy Q generated per gm of   ${}_{92}{}^{235}U$ = 200$*$2.563 $*{10}^{21}$ MeV/g = 5.126$*{10}^{23}$ 
 MeV/g = 5.126 $*{10}^{23}*1.6*{10}^{-19}*{10}^6$ J/g = 8.20$*{10}^{10}$ 
 J/g

Since the reactor operates only 80% of the time, hence the amount of ${}_{92}{}^{235}\mathit{U}$
 in 5 years is given by $\frac{0.8*157.68*{10}^6*1000*{10}^6}{8.20*{10}^{10}}$ = 1538 kg

Hence, initial amount of fuel = 2 $*1538kg$ = 3076 kg

13.17 The average energy released per fission of ${}_{94}{}^{239}Pu$  $E_{avg}$= 180 MeV

Amount of pure${}_{94}{}^{239}u$ , m = 1 kg = 1000 g

Avogadro's number,
$N_A$ = 6.023 $*{10}^{23}$

Mass number of ${}_{94}{}^{239}\mathit{P}\mathit{u}$  = 239 gm

Hence, number of atoms in 1000 g ${}_{94}{}^{239}Pu$, N =$\frac{6.023*{10}^{23}}{239}$ $*1000$= 2.52 $*{10}^{24}$

Total energy released during the fission of 1 kg of  ${}_{94}{}^{239}Pu$, E =$E_{avg}$ $*$N

= 180 $*$ 2.52$*{10}^{24}$ MeV = 4.536 $*{10}^{26}$ MeV

13.16 The fission can be shown as:

${}_{26}{}^{56}Fe\to 2{}_{13}{}^{28}Al$ 

It is given that atomic mass

m ( ${}_{26}{}^{56}Fe)$= 55.93494 u

m ( ${}_{13}{}^{28}Al)$= 27.98191 u

The Q-value of this reaction is given as:

Q = $\left[m\left({}_{26}{}^{56}Fe\right)-2m\left({}_{13}{}^{28}Al\right)\right]c^2$

= $\left[55.93494\mathrm{}-2*27.98191\right]c^2$

= -0.02888 $c^2$

= -0.02888$*931.5$  MeV

= - 26.902 MeV

The Q value of the fission is negative, therefore the fission is not possible energetically. Q value needs to be positive for a fission.

13.11 Nuclear radius of the gold isotope,  ${}_{47}{}^{197}Au$ = $R_{Au}$

Nuclear radius of silver isotope,  ${}_{47}{}^{107}Ag$ = $R_{Ag}$

Mass number of gold,  $A_{Au}$ = 197

Mass number of silver,  $A_{Ag}$ = 107

The ratio of the radii of the two nuclei is related with their mass numbers as :

$\frac{R_{Au}}{R_{Ag}}$ = ${\left. [\frac{A_{Au}}{A_{Ag}}\right]}^{1/3}$ = ${\left. [\frac{197}{107}\right]}^{1/3}$ =1.2256

Hence, the ratio of the nuclear radii of the gold and silver isotope is 1.23