Physics Ncert Solutions Class 12th

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

13.10 Half life of ${}_{38}{}^{90}Sr$ , $T_{1/2}$ = 28 years = 28 $*365*24*60*60$ secs = 0.883 $*{10}^9$ s

Mass of the isotope, m = 15 mg = 15 $*{10}^{-3}$ gms

90 g of ${}_{38}{}^{90}Sr$ contains 6.023 $*{10}^{23}$ atoms

No. of atoms in 15 mg of ${}_{38}{}^{90}Sr$ contains = $\frac{6.023*{10}^{23}}{90}*$ 15 $*{10}^{-3}$ = 1.0038 $*{10}^{20}$

Rate of disintegration $\frac{dN}{dt}$ = $?N$ , where $?$ = $\frac{0.693}{T_{1/2}}$ = $\frac{0.693}{0.883*{10}^9}$ /s = 7.848 $*{10}^{-10}$ $s^{-1}$

$\frac{dN}{dt}$ = 7.848 $*{10}^{-10}*$ 1.0038 $*{10}^{20}$ = 7.878 $*{10}^{10}$ atoms / second.

13.9 The strength of the radioactive source is given as:

$\frac{dN}{dt}$ = 8.0 mCi = 8 $*{10}^{-3}$ $*3.7*{10}^{10}$ decay/s = 296 $*{10}^6$ decay/s, where

N = Required number of atoms

Given, half life of ${}_{27}{}^{60}Co$ , $T_{1/2}$ = 5.3 years = 5.3 $*365*24*60*60$ secs = 167 $*{10}^6$ s

For decay constant $\lambda$ , we have rate of decay as:

$\frac{dN}{dt}$ = $\lambda N$ or

N = $\frac{1}{\lambda }\frac{dN}{dt}$ , where $\lambda$ = $\frac{0.693}{T_{1/2}}$ = $\frac{0.693}{167\mathrm{}*{10}^6}$ /s = 4.1497 $*{10}^{-9}$ $s^{-1}$

N = $\frac{296\mathrm{}*{10}^6\mathrm{}}{4.1497*{10}^{-9}}$ = 7.133 $*{10}^{16}$ atoms

For ${}_{27}{}^{60}Co$ , mass of 6.023 $*{10}^{23}$ atoms = 60 gms

Therefore, the mass of 7.133 $*{10}^{16}$ atoms = $\frac{60}{6.023*{10}^{23}}*$ 7.133 $*{10}^{16}$ gms = 7.106 $*{10}^{-6}$ g