Draw the graph to show variation of binding energy per nucleon with mass number of different atomic nuclei. Calculate binding energy/nucleon of ${}_{20}^{40}{Ca}$ nucleus.
Given:
mass of ${}_{20}^{40}{Ca} = 39.962589 u$
mass of proton = 1.007825 u
mass of neutron = 1.008665 u
and 1 u = 931 MeV/c^2.

Solution

The graph showing the variation of binding energy per nucleon with mass number of atomic nuclei is shown on below.

Numerical:

Q-value for the gien nucleus is,

Q = [20 m_{n} + 20 m_{p} - m (Ca)] c^{2}

= (20 \times 1.008665 + 20 \times 1.007825 - 39.962589) c^{2}

= 0.367211 u \times \frac{931.5}{u} MeV = 342 MeV

Hence, the B.E. per nucleon of

{}_{20}^{40}{Ca}

nucleus is,

\frac{Q}{A} = \frac{342}{40} = 8.55 MeV / nucleon .

For Daily Free Study Material Join wiredfaculty