Question

How do you calculate binding energy in $\mathrm{MeV}$?

Answer 1

• The following equation gives binding energy for nucleus:

$BE=\left\{\right[Zm\left({}^{1}H\right)+Nmn]-m({}^{\mathrm{A}}X\left)\right\}{c}^{\mathit{2}}$

• For ${}^{\mathit{4}}He$, Substituting the above values in the equation we got:
• Since, $$\begin{gathered} Z=N \\ Z=2 \end{gathered}$$ thus, $BE=\left[2m\right({}^{1}H)+2mn]-m\left({}^{4}He\right)c^{\mathit{2}}.$
• We know the following values:

$$\begin{gathered} m\left({}^{4}He\right)=4.002602u,m \\ \left({}^{1}H\right)=1.007825u,mn=1.008665u. \\ \mathrm{BE}=(0.030378u)c^{\mathit{2}} \end{gathered}$$

• Noting that $1\mathrm{u}=935.15\raisebox{1ex}{$\mathrm{MeV}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{c}}^2$}\right.$
• Substituting the above values in the equation we get:

$$\begin{gathered} \mathrm{BE}=(0.030378)(931.5MeV/c^{\mathit{2}})c \\ BE\mathit{}=28.3MeV. \end{gathered}$$

• Since $\mathrm{A}=4$, we further simplify as follows:

$\raisebox{1ex}{$\mathrm{BE}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.=7.07\raisebox{1ex}{$\mathrm{MeV}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{nucleon}.$}\right.$

The above points clearly explains the calculation of binding energy.