Class 12th

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

The mass defect of the reaction is given as:

Δm = m ${}_{20}{}^{40}Ca)+$ (  m( ${}_0{}^{1}n)-$ m( ${}_{20}{}^{41}Ca)$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $*{10}^3$ u

=8.363007 MeV

For ${}_{13}{}^{27}Al$ , the neutron removal reaction can be written as

${}_{13}{}^{27}Al\to$ ${}_{13}{}^{26}Al$ + ${}_0{}^{1}n$

It is given that

m( ${}_{13}{}^{26}Al)$ = 25.986895 u

m( ${}_{13}{}^{27}Al)$  = 26.981541 u

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

The mass defect of the reaction is given as:

Δm = m( ${}_{13}{}^{26}Al)+$  m( ${}_0{}^{1}n)-$ m( ${}_{13}{}^{27}Al)$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

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

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

= (-4.33 $*{10}^{-3}$ ) $c^2$

But 1 u = 931.5 MeV/ $c^2$

Q = -4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

The given nuclear reaction is

${}_6{}^{12}C$ + ${}_6{}^{12}C$ $\to$ ${}_{10}{}^{20}Ne$ + ${}_2{}^{4}He$

Atomic mass

m ( ${}_6{}^{12}C$ ) = 12.000000 u

m ( ${}_{10}{}^{20}Ne$ ) = 19.992439 u

m ( ${}_2{}^{4}He$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $\left[2m\left({}_6{}^{12}C\right)-\mathrm{}m\left({}_{10}{}^{20}Ne\right)-m\left({}_2{}^{4}He\right)\mathrm{}\right]c^2$

= $\left[2*12.0-19.992439-4.002603\right]c^2$

=4.958 $*{10}^{-3}{c}^2$ u

=4.958 $*{10}^{-3}*931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

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