Question

The mass of ${}_3^{7}Li$ is $0.042$ less than the mass of $3$ protons and $4$ neutrons. The binding energy per nucleon in ${}_3^{7}Li$ is:

• A. $5.6MeV$
• B. $56MeV$
• C. $0.56MeV$
• D. $560MeV$

Answer 1

Answer: (A)

$5.6MeV$

The mass defect for ${}_3^{7}Li$ is $\Delta m=0.042$ amu.
1 amu $=931.5$ MeV
Binding energy $=0.042\times 931.5=39.123$ MeV
Total number of nucleons in ${}_3^{7}Li$ is 7 (3 protons and 4 neutrons).
Binding energy per nucleon $=\frac{39.123}{7}=5.6$ MeV.