Question

Calculate $\mathbf{Q}$ for neutron decay.

Short answer

The value of $Q$ is $0.782\text{\hspace{0.17em}MeV}$.

Step-by-step solution

Given data

Neutron decay reaction is ${}_0^1\text{n}{\to }_1^1\text{p}{+}_{-1}^0\mathrm{e}$.

Atomic mass of${}_0^1\text{n},{\mathrm{m}}_{\mathrm{n}}=1.008665\text{u}$.

Atomic mass of${}_1^1\text{P},{\mathrm{m}}_{\mathrm{p}}=1.007828\text{u}$ .

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 value of Q  for neutron decay

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

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

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\mathrm{n}}-{\mathrm{m}}_{\mathrm{p}})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=(1.008665\text{u}-1.007825\text{u})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=0.782\text{MeV}\end{array}$

The value of $Q$ is $0.782\text{\hspace{0.17em}MeV}$.