The binding anergy per nucleon is `8.5 MeV` for `A=120` and is `7.6 MeV` for `A=240` (see in figure). Suppose a nucleus with `A=240` breaks into two nuclei of nearly equal mass numbers. Calculate the energy released in the process.

Explanation

Suppose the heavy nucleus had `Z` protons and `N` neutrons. The rest mass energy of this nucleus would be
`E = Mc^2 = (Zm_p + Nm_n) C62 - B_1`
`= (Zm_p + Nm_n) c^2 - 7.6 xx 240 MeV`.
If there are `Z_1` protons and `N_1` neutrons in the first fragment, its rest mass energy will be
`E_1 = M_1 c^2 = (Z_1 m_p + N_1 m_n) c^2 - B_2`
`= (Z_1 m_p + N_1 m_n)c^2 - (8.5 MeV) (Z_1 + N_1)`.
Similarly, if there are `Z_2` protons and `N_2` neutrons in the second fragment, its rest mass energy will be
`E_2 = (Z_2 m_p + N_2 m_n) c^2 - (8.5 MeV) (Z_2 + N_2)`.
The energy released due to the breaking is
`E - (E_1 + E_2)`.
`=[(Z-Z_1 - Z_2) m_p c^2 + (N - N_1 - N_2) m_n c^2]`.
`+ [(Z_1 + Z_2 + N_1 + N_2) xx 8.5 - 240 xx 7.6] MeV`
`=240 xx (8.5 - 7.6) MeV = 216 MeV`.
We have used the fact that `Z_1 + Z_2 = Z, N_1 + N_2 = N` and `Z_1 + Z_2 + N_1 + N_2 = Z + N = 240`. Thus, `216, `216 MeV` of energy will be released when this nucleus breaks.