Class 12th

10.1 Wavelength of incident monochromatic light, $?$ = 589 nm = 589 $*{10}^{-9}$ m

Speed of light in air, c = 3 $*{10}^8$ m/s

Refractive index of water, $?$ = 1.33

In case of reflection, the ray goes back to the same medium. Hence wavelength, frequency and speed of reflected beam will be same as incident beam.

Frequency of light beam is given by the relation, $?$ = $\frac{c}{?}$ = $\frac{3\mathrm{}*{10}^8}{589\mathrm{}*{10}^{-9}}$ = 5.09 $*{10}^{14}$ Hz.

Hence speed = 3 $*{10}^8$ m/s, Wavelength = 589 $*{10}^{-9}$ m, Frequency = 5.09 $*{10}^{14}$ Hz of incident ray and reflected ray will remain unchanged.

(i) Frequency = does not depend on the property of the medium in which it is travelling, hence frequency will remain same, i.e. 5.09 $*{10}^{14}$ Hz

(ii) Speed of light in water depends upon the refractive index of water, hence speed of light, v = $\frac{c}{?}$ = $\frac{3\mathrm{}*{10}^8}{1.33}$ = 0.226 $*{10}^9$ m/s

Wavelength in water, $?$ = $\frac{v}{?}$ = $\frac{0.226\mathrm{}*{10}^9}{5.09\mathrm{}*{10}^{14}}$ = 4.432 $*{10}^{-7}$ m = 443.2 nm

Hence the speed, frequency and wavelength of refracted light are 0.226 $*{10}^9$ m/s, 5.09 $*{10}^{14}$ Hz and 443.2 nm

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