91Ó°ÊÓ

The reaction energy is determined by using the expression:

$\mathbf{Q}\mathbf{=}\left[M_A+M_B-M_C-M_D\right]\mathbf{×}{\mathbf{c}}^{\mathbf{2}}$

Where, ${\mathit{M}}_{\mathbf{A}}$and ${\mathit{M}}_{\mathbf{B}}$are the mass of reactants, ${\mathit{M}}_{\mathbf{c}}$and ${\mathit{M}}_{\mathbf{D}}$are the mass of products.

$\mathit{c}$is the speed of light that and ${\mathit{c}}^{\mathbf{2}}\mathbf{=}\mathbf{931}\mathbf{.}\mathbf{5}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u}$.

The given reaction is:

${\mathrm{e}}^{-}+p→n+v_{\mathrm{e}}$

The atomic mass of e is $0.000548580u$, the atomic mass of $\mathrm{ÒÏ}$is $1.007276u$, the atomic mass of $n$is $1.008665u$and the atomic mass of $v$is 0 .

Now,

localid="1667794539305" $Q=\left[M\left(e^{-}\right)+M\left(p\right)-M\left(n\right)-M\left(v_e\right)\right]$

Substituting the values, and we get:

localid="1663937569164" $\begin{array}{rcl}Q& =& \left[\right(0.000548580)+(1.007276)-(1.008665)-(0\left)\right](931.5)\\ & =& (-0.000840420)(931.5)\\ & =& -0.783\mathrm{MeV}\end{array}$

As the reaction energy $Q$is negative, so the reaction is endoergic.

Hence, the reaction energy Q for the reaction localid="1667794578268" $e^{-}+\mathrm{ÒÏ}→n+v_e$is $-0.783\mathrm{MeV}$, and the reaction is endoergic.