Question

How much Kinetic energy released and what is the daughter nucleus in the ${\mathbf{\beta }}^{\mathbf{+}}$ decay of nitrogen-13?

Short answer

Kinetic energy release in ${\mathrm{\beta }}^{+}$decay is$1.708\text{\hspace{0.17em}MeV}$ and daughter nucleus is Carbon-13.

Step-by-step solution

Given data

The ${\mathrm{\beta }}^{+}$decay of nitrogen- .

role="math" localid="1658409778492" ${}_7^{13}\text{N}{\to }_6^{13}\text{C}{+}_{+1}^0{\mathrm{\beta }}^{+}+\mathrm{v}$

Atomic mass of ${}_7^{13}\text{N},{\mathrm{m}}_{\text{N}}=13.005738\text{u}$.

Atomic mass of ${}_6^{13}\mathrm{C},{\mathrm{m}}_{\mathrm{C}}=13.003355\text{u}$.

Atomic mass of electron,${\mathrm{m}}_{\mathrm{e}}=5.486\times {10}^{-4}\text{u}$ .

Definition of Kinetic Energy 

The kinetic energy released is, $\mathbf{Q}\mathbf{=}\mathbf{(}{\mathbf{m}}_{\mathbf{i}}\mathbf{-}{\mathbf{m}}_{\mathbf{f}}\mathbf{)}{\mathbf{c}}^{\mathbf{2}}$.

Where,${\mathbf{m}}_{\mathbf{i}}$ is the mass of parents’ nucleus, ${\mathbf{m}}_{\mathbf{f}}$is the mass of the daughter nucleus, and $\mathbf{c}$is the speed of light in vacuum.

Determine the Kinetic Energy released

Kinetic energy release in ${\mathrm{\beta }}^{+}$decay is $\mathrm{Q}=({\mathrm{m}}_{\text{parent}}-{\mathrm{m}}_{\text{daughter}}-{\mathrm{m}}_{\mathrm{e}}){\mathrm{c}}^2$.

The conversion factor may be used to express, ${\mathrm{c}}^2\to \text{MeV}/\text{u}$ .

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{purent}}-{\mathrm{m}}_{\text{daughter}}-{\mathrm{m}}_{\mathrm{e}})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=({\mathrm{m}}_{\text{N}}-{\mathrm{m}}_{\mathrm{C}}-2{\mathrm{m}}_{\mathrm{e}})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=[13.005738\text{u}-13.003355\text{u}-5.486\times {10}^{-4}\text{u}]\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=1.708\text{MeV}\end{array}$

Kinetic energy release in ${\mathrm{\beta }}^{+}$decay is $1.708\text{\hspace{0.17em}MeV}$ and daughter nucleus is Carbon- $13$.