Calculate (in $MeV$) the binding energy per nucleon for $H{e}^2$and $H{e}^4$. Which is more tightly bound?

We have been given that,

$H{e}^2$, $H{e}^4$

We need to find the binding energy per nucleon for both atoms.

The binding energy is given by, $H{e}^2,Z=2,N=2,A=2$

$BE=\left(Z{m}_P+N{m}_n-m_A\right)MeV$

Where,

$m_p$=mass of proton ,$m_n$=mass of neutron, $m_A$= mass of atom

$$\begin{gathered} BE=\left(2\times 1.0087u+0\times 1.0078u-4.0026\right)931.49MeV/u \\ =(2.0174-4.0026)931.49MeV \\ =-1849.1939MeV \end{gathered}$$

Now,

$BE=-1849.1939MeV,A=2$

The binding energy per nucleon is = $\frac{BE}{A}$

$$\begin{gathered} =\frac{-1849.1939}{2}MeV \\ =-924.5969MeV \end{gathered}$$

Binding energy is given by ,$H{e}^4,Z=2,A=4,N=2$

$BE=\left(Z{m}_P+N{m}_N-m_A\right)MeV$

$$\begin{gathered} BE=\left(2\times 1.0087u+2\times 1.0078u-4.0026\right)931.49MeV/u \\ =0.0304\times 931.49MeV \\ =28.317MeV \end{gathered}$$

Now,$BE=28.317MeV,A=4$

Binding energy per nucleon is = $\frac{BE}{A}MeV$

= $\frac{28.317}{4}MeV$

= $7.079MeV$

Since the binding energy per nucleon :$H{e}^4>H{e}^2$

Therefore, $H{e}^4$is more tightly bounded.