Question

If the binding energy per nucleon for ${}_3{Li}^7$ is $5.6MeV$, the total binding energy of a lithium nucleus is?

• A. $139.2MeV$
• B. $39.2MeV$
• C. $15.8MeV$
• D. $115.8MeV$

Answer 1

The correct option is B $39.2MeV$

Binding energy per nucleon = 5.6 MeV

No. of nucleon = No. of proton + No. of neutron

=3 + 4 = 7

So, for 7 Nucleon $=7\times 5.6=39.2MeV$