Nuclei

X + p → Y + b

$Q=K_Y+K_b-K_P$

$Q+K_P=K_Y+K_b$                

$\Rightarrow Q+K_P>0$                 

For angle of incidence ‘i’ :-

cos I = $\frac{5}{\sqrt[]{48+27+25}}=\frac{5}{\sqrt[]{100}}$

i = 60°

Using snell’s law :-

${\mu }_{1}sini={\mu }_{2}sinr$

sin r = $\frac{\sqrt[]{2}}{\sqrt[]{3}}sin60°=\sqrt[]{\frac{2}{3}}\times \frac{\sqrt[]{3}}{2}=\frac{1}{\sqrt[]{2}}$

$\therefore r=45°$

So, difference, I – r = 60° - 452 = 15°

Sol. $\Rightarrow \mathrm{}{\stackrel{?}{\mathrm{V}}}_1=2\stackrel{?}{{\mathrm{U}}_{\mathrm{c}\mathrm{m}}}-\stackrel{?}{{\mathrm{U}}_1}\Rightarrow \mathrm{}2\left[\frac{\mathrm{m}/2{\mathrm{V}}_{\mathrm{o}}}{\frac{\mathrm{m}}{2}+\frac{\mathrm{m}}{3}}\right]-{\mathrm{V}}_{\mathrm{o}}$

$\Rightarrow \mathrm{}\frac{6}{5}{\mathrm{V}}_{\mathrm{o}}-{\mathrm{V}}_{\mathrm{o}}\Rightarrow \frac{{\mathrm{V}}_0}{5}$

$\Rightarrow \mathrm{}{\lambda }_{\mathrm{o}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{m}}{2}{\mathrm{V}}_{\mathrm{o}}\right)}\mathrm{}{\lambda }_{\mathrm{f}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{M}}{2}\frac{{\mathrm{V}}_{\mathrm{o}}}{5}\right)}$

$\Rightarrow \mathrm{\Delta }\lambda =\frac{8\mathrm{h}\mathrm{c}}{{\mathrm{m}\mathrm{V}}_{\mathrm{o}}}$

13.28 The equation for deuterium-tritium fusion is given as:

${}_1{}^{2}H+{}_1{}^{3}H\to {}_2{}^{4}He+n$

 It is given that

Mass of ( ${}_1{}^2\mathrm{H}),\mathrm{}{\mathrm{m}}_1$  = 2.014102 u

Mass of ( ${}_1{}^3\mathrm{H})$ , ${\mathrm{m}}_2=\mathrm{}$ 3.016049 u

Mass of ( ${}_2{}^{4}He),m_3$  = 4.002603 u

Mass of ( ${}_0{}^{1}n),m_4$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $\left[{\mathrm{m}}_1+{\mathrm{m}}_2-m_3-m_4\right]c^2$

= $\left[2.014102+3.016049-4.002603-1.008665\right]c^{2}u$

= 0.018883 $c^{2}u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0fm$

13.27 In the fission of ${}_{92}{}^{238}U$ , ${\beta }^{-}$ 10  particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}{}^{238}U+{}_0{}^{1}n\to {}_{58}{}^{140}Ce+{}_{44}{}^{99}Ru+10{}_{-1}{}^{0}e$

It is given that:

Mass of a ${}_{92}{}^{238}\mathrm{U}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_{1\mathrm{}}$ = 238.05079 u

Mass of a ${}_{58}{}^{140}\mathrm{C}\mathrm{e}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_2$ =139.90543 u

Mass of a ${}_{44}{}^{99}\mathrm{R}\mathrm{u}\mathrm{}\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s},\mathrm{}{\mathrm{m}}_3$ = 98.90594 u

Mass of a neutron ${}_0{}^{1}n$ , $m_4$ = 1.00865 u

Q value of the above equation,

Q = $\left[m^{\text{'}}\left({}_{92}{}^{238}\mathrm{U}\right)+\mathrm{}\mathrm{m}\mathrm{\text{'}}({}_0{}^1\mathrm{n})-{\mathrm{m}}^{\mathrm{\text{'}}}({}_{58}{}^{140}Ce)-m^{\text{'}}({}_{44}{}^{99}Ru)-10m_e\right]c^2$

Where

m’ = re

13.26 For the emission of ${}_6{}^{14}C$ , the nuclear reaction is:

${}_{88}{}^{223}Ra\to {}_{82}{}^{209}Pb+{}_6{}^{14}C$

We know that:

Mass of ${}_{88}{}^{223}Ra$ , $m_1$ = 223.01850 u

Mass of ${}_{82}{}^{209}Pb$ , $m_2$  = 208.98107 u

Mass of ${}_6{}^{14}C$ , $m_3$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_1$ - $m_2$  - $m_3$ ) $c^2$

= (223.01850 - 208.98107 - 14.00324) $c^2$ u

= 0.03419 $c^2$ u

= 0.03419 MeV = 3

13.24 If a neutron ${}_0{}^{1}n)$  is removed from ${}_{20}{}^{41}Ca$ , the corresponding reaction can be written as:

${}_{20}{}^{41}Ca\to$ ${}_{20}{}^{40}Ca$ + ${}_0{}^{1}n$

The separation energies are

For ${}_{20}{}^{41}Ca$  : Separation energy = 8.363007 MeV

For ${}_{13}{}^{27}Al$  : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}{}^{40}Ca)$  = 39.962591 u

m( ${}_{20}{}^{41}Ca)$ = 40.962278 u

m( ${}_0{}^{1}n)$ = 1.008665 u

Th

13.15 The given nuclear reaction is

${}_1{}^{1}H$ + ${}_1{}^{3}H$ $\to$ ${}_1{}^{2}H$ + ${}_1{}^{2}H$

Atomic mass

m ( ${}_1{}^{1}H$ ) = 1.007825 u

m ( ${}_1{}^{2}H$ ) = 2.014102 u

m ( ${}_1{}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $\left[m\left({}_1{}^{1}H\right)+\mathrm{}m\left({}_1{}^{3}H\right)-\mathrm{}2m\left({}_1{}^{2}H\right)\right]c^2$

= $\left[1.007825+\mathrm{}3.016049-2\times 2.014102\right]c^2$

= (-4.33 $\times {10}^{-3}$ ) $c^2$

But 1 u = 931.5 MeV/ $c^2$

Q = -4.0334 Me

13.13 The given values are

m ( ${}_6{}^{11}C)$ = 11.011434 u and m ( ${}_5{}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_6{}^{11}C\to {}_5{}^{11}B+e^{+}+v$

Half life of ${}_6{}^{11}C$ nuclei, $T_{1/2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_6{}^{11}C$

ΔQ

13.12$\alpha -$ particle decay of ${}_{88}{}^{226}Ra$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}{}^{226}Ra$ $\to$ ${}_{86}{}^{222}Rn$ + ${}_2{}^{4}He$

Q value of emitted $\alpha -$ particle = (Sum of initial mass – Sum of final mass) $\times c^2$ , where

c = Speed of li