Physics Ncert Solutions Class 12th

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 = $\left[m\text{'}({}_6{}^{11}C)-\{m\text{'}\left({}_5{}^{11}B\right)+{\mathrm{m}}_{\mathrm{e}}\}\mathrm{}\right]c^2\dots \dots \dots \dots \dots \left(1\right)$

where

$m_e$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m' = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_e$ in the case of ${}_6{}^{11}C$ and 5 $m_e$ in the case of ${}_5{}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $\left[m({}_6{}^{11}C)-m\left({}_5{}^{11}B\right)-{2\mathrm{m}}_{\mathrm{e}}\right]c^2$

= $\left[11.011434-11.009305-2*0.000548\right]c^2$ u

=1.033 $*{10}^{-3}{c}^2$ u

But 1 u = 931.5 MeV/ $c^2$

ΔQ = 0.962 MeV

The value of $\Delta Q$ is almost comparable to the maximum energy of the emitted positron.

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) $*c^2$ , where

c = Speed of light.

It is given that

m ( ${}_{88}{}^{226}Ra)$ = 226.02540 u

m ( ${}_{86}{}^{222}Rn)$ = 222.01750 u

m ( ${}_2{}^{4}He)$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^2$

= 5.297 $*{10}^{-3}{uc}^2$

But 1 u = 931.5 MeV/ $c^2$

Hence Q = 4.934 MeV

Kinetic energy of the $\alpha -$ particle = $\frac{Massnumberafterdecay}{Massnumberbeforedecay}$ $*Q$ = $\frac{222}{226}$ $*$ 4.934= 4.85 MeV

$\alpha -$ particle decay of ${}_{86}{}^{220}Rn$ emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.

${}_{86}{}^{220}Rn$ $\to$ ${}_{84}{}^{216}Po$ + ${}_2{}^{4}He$

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

c = Speed of light.

It is given that

m ( ${}_{86}{}^{222}Rn)$ = 220.01137 u

m ( ${}_{84}{}^{216}Po)$ = 216.00189 u

m ( ${}_2{}^{4}He)$ = 4.002603 u

Q value = [(220.01137) – (216.00189 + 4.002603)] $c^2$

= 6.877 $*{10}^{-3}{uc}^2$

But 1 u = 931.5 MeV/ $c^2$

Hence Q = 6.406 MeV

Kinetic energy of the $\alpha -$ particle = $\frac{Massnumberafterdecay}{Massnumberbeforedecay}$ $*Q$ = $\frac{216}{222}$ $*$ 6.406= 6.23 MeV

13.1 Mass of ${}_3{}^{6}Li$ lithium isotope, $m_1$ = 6.01512 u

Mass of ${}_3{}^{7}Li$ lithium isotope, $m_2$ = 7.01600 u

Abundance of ${}_3{}^{6}Li$ , ${\eta }_1$ = 7.5%

Abundance of ${}_3{}^{7}Li$ , ${\eta }_2$ = 92.5%

The atomic mass of lithium atom is given as:

m = $\frac{m_1{\eta }_1+m_{2{\eta }_2}}{{\eta }_1+{\eta }_2}$ = $\frac{6.01512\mathrm{}*7.5+\mathrm{}7.01600\mathrm{}*92.5}{7.5+92.5}$ = 6.940934 u

Mass of ${}_5{}^{10}B$ Boron isotope, $m_1$ = 10.01294 u

Mass of ${}_5{}^{11}B$ Boron isotope, $m_2$ = 11.00931 u

Let the abundance of ${}_5{}^{10}B$ be x % and that of ${}_5{}^{11}B$ be (100-x) %

The atomic mass of Boron atom is given as :

10.8111 = $\frac{10.01294x+11.00931(100-x)}{x+(100-x)}$

1081.11 = 1100.931 - 0.99637x

x = 19.89 %

Hence the abundance of ${}_5{}^{10}B$ is 19.89 % and that of ${}_5{}^{11}B$ is (100-19.89) = 80.11 %