91Ó°ÊÓ

We have been given that,

$H^3$,$H{e}^3$

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

Binding energy is given by,

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

Where,

$Z$= atomic number ,$N$= number of neutrons ,$m_A=$mass of atom,

$m_N$=mass of neutron , $m_E$=mass of element.

For $H^3$atom, $$\begin{gathered} BE=\left(1×1.00783u+2×1.00866u-3.01605u\right)×931.49MeV/u \\ =(3.02515-3.01605)×931.49MeV/u \\ =8.48MeV \end{gathered}$$

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

Therefore,

For $H^3$, $BE=8.48MeV,A=3$

Binding energy per nucleon = $\frac{8.48}{3}MeV$

= $2.83MeV$

Binding energy is given by,

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

For $H{e}^3$,

$Z=1,N=2,A=3$

$$\begin{gathered} BE=\left(1×1.00783u+2×1.783u-3.01605u\right)931.49MeV/u \\ =\left(3.02515-3.0164\right)931.49MeV \\ =7.78MeV \end{gathered}$$

Now,

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

= $\frac{7.78}{3}MeV$ $\left(A=3\right)$

= $2.594MeV$