Physics

13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $*{10}^5$  MW = 2 $*{10}^4$ MW = 2 $*{10}^4*{10}^6$  J/s

= 2 $*{10}^4*{10}^6*60*60*24*365$  J/y = 6.3072 $*{10}^{17}$ J/y

Heat energy release per fission of a ${}_{}{}^{235}U$  nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $*{10}^6$ eV = 50 $*{10}^6*1.6*{10}^{-19}$ J = 8 $*{10}^{-12}$  J

Therefore, number of atoms required per year = $\frac{6.3072*{10}^{17}}{8*{10}^{-12}}$  = 7.884 $*{10}^{28}$

1 mole of ${}_{}{}^{235}U$  = 235 gm of ${}_{}{}^{235}U$  contains 6.023 $*{10}^{23}$ atoms

Hence the mass of 7.884 $*{10}^{28}atoms$  =$\frac{235*{10}^{-3}}{6.023*{10}^{23}}$ $*$ 7.884 $*{10}^{28}$ = 30.76 $*{10}^3$ kg

Hence, the Uranium needed per year is 30.76$*{10}^3$ kg

13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_1{}^{1}H)$ contains 6.023 $*{10}^{23}$  atoms

1 kg of hydrogen contains = 1000 $*$ 6.023 $*{10}^{23}$  atoms = 6.023 $*{10}^{26}$ atoms

Within Sun, four ( ${}_1{}^{1}H)$  nuclei combine and forms 1 ${}_2{}^{4}He$  nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_1{}^{1}H$  is:

$E_1$  = $\frac{6.023*{10}^{26}*26}{4}$  MeV = 3.91495 $*{10}^{27}$ MeV

Amount of ${}_{92}{}^{235}U$ , m = 1 kg = 1000 g

1 mole of ${}_{92}{}^{235}U$ , i.e. 235 g of ${}_{92}{}^{235}U$  contains 6.023 $*{10}^{23}$ atoms

1 kg of  ${}_{92}{}^{235}U$ contains = $\frac{1000}{235}$ $*$  6.023 $*{10}^{23}$ atoms = 2.563 $*{10}^{24}$  atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}{}^{235}U$  is 200 MeV.

Hence the energy released from fusion of 1 kg of  ${}_{92}{}^{235}U$  is:

 = 2.563 $*{10}^{24}*200$  MeV = 5.126 $*{10}^{26}$ MeV

$\frac{E_1}{E_2}$ = $\frac{3.91495*{10}^{27}}{5.126*{10}^{26}}$ = 7.64

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of ${}_{92}{}^{235}U$ .

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 $P_{15}^{33}$