Question

Calculate the energy released in the${\beta }^{+}$decay of${}^{22}\mathrm{N}\mathrm{a}$, the equation for which is given in the text. The masses of${}^{22}\mathrm{N}\mathrm{a}$and${}^{22}\mathrm{N}\mathrm{e}$are$21.994434$and$21.991383\mathrm{u}$, respectively.

Short answer

The energy released is$1.820\mathrm{M}\mathrm{e}\mathrm{V}$.

Step-by-step solution

Define energy

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

Explanation

The${}^{22}\mathrm{N}\mathrm{a}$having${\beta }^{+}$decay is,

${}_{11}^{22}N{a}_{11}\underset{{\beta }^{+}}{\overset{}{\to }}{}_{10}^{22}N{e}_{12}+{\beta }^{+}+v_e$

The energy can be calculated by,

$\begin{array}{rl}E& =\mathrm{\Delta }m{c}^2\\ & =(M_{{22}_{Na}}-M_{{22}_{Ne}}-2m_e)c^2\\ & =\left[(21.994434\mathrm{u}-21.991383\mathrm{u}-2(0.00054858\mathrm{u}))931.5\mathrm{M}\mathrm{e}\mathrm{V}/\mathrm{u}{\mathrm{c}}^2\right]c^2\\ & =1.820\mathrm{M}\mathrm{e}\mathrm{V}\end{array}$

Therefore, the energy released is $1.820\mathrm{M}\mathrm{e}\mathrm{V}$.