Q-39PE - College Physics (Urone)

Question

Calculate the energy released in the \({{\rm{\beta }}^{\rm{ + }}}\) decay of \({}^{{\rm{22}}}{\rm{Na}}\) , the equation for which is given in the text. The masses of \({}^{{\rm{22}}}{\rm{Na}}\) and \({}^{{\rm{22}}}{\rm{Ne}}\) are \({\rm{21}}{\rm{.994434}}\) and\({\rm{21}}{\rm{.991383u}}\), respectively.

Step-by-Step Solution

Answer

The energy released is\({\rm{1}}{\rm{.820MeV}}\).

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 \({}^{{\rm{22}}}{\rm{Na}}\) having \({{\rm{\beta }}^{\rm{ + }}}\) decay is,

 \({}_{11}^{22}N{a_{11}}\xrightarrow[{{\beta ^ + }}]{}{\underline{\phantom{xx}}}_{10}^{22}N{e_{12}} + {\beta ^ + } + {v_e}\)

The energy can be calculated by,

\(\begin{align}{\underline{\phantom{xx}}}E &= \Delta m{c^2}\\ & = ({M_{{{22}_{Na}}}} - {M_{{{22}_{Ne}}}} - 2{m_e}){c^2}\\ & = \left[ {(21.994434\,{\rm{u}} - 21.991383\,{\rm{u}} - 2(0.00054858\,{\rm{u}})){\rm{931}}{\rm{.5}}\,{\rm{MeV/u}}{{\rm{c}}^{\rm{2}}}} \right]{c^2}\\ & = 1.820\,{\rm{MeV}}\end{align}\)

 

Therefore, the energy released is \(1.820\,{\rm{MeV}}\).