Question

Find Q for the decay of beryllium-10

Short answer

The $\mathrm{Q}$ factor is $0.56\text{\hspace{0.17em}MeV}$.

Step-by-step solution

Given data

Given that,${}_4^{10}\text{Be}{\to }_5^{10}\text{B}{+}_{-1}^0\text{e}+\overline{\mathrm{v}}$.

Atomic mass of ${}_4^{10}\text{Be},m_{\text{Be}}=10.013534\text{u}$.

Atomic mass of ${}_5^{10}\text{B},{\mathrm{m}}_{\text{B}}=10.012937\text{u}$.

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

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

$\begin{array}{c}\mathrm{Q}=({\mathrm{m}}_{\text{Be}}-{\mathrm{m}}_{\text{B}})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=(10.013534\text{u}-10.012937\text{u})\left(931.5\frac{\text{MeV}}{\text{u}}\right)\\ \mathrm{Q}=0.56\text{MeV}\end{array}$

The $\mathrm{Q}$ factor is $0.56\text{\hspace{0.17em}MeV}$.