Question

Find the binding energy per nucleon for lithium nucleus ${}_3{Li}^7$.

Take mass of ${}_3{Li}^7=7u$, mass of proton $=1.007825u$ and mass of neutron $=1.008665u$.

Take $1u=931.5MeV$.

• A. $3.7MeV$
• B. $4.9MeV$
• C. $1.7MeV$
• D. $7.736MeV$

Answer 1

The correct option is D $7.736MeV$

Binding energy per nucleon. $E=\frac{△E}{A}=\frac{△m\ast 931}{A}$

$△m=(3M_p+4M_n)-$ Mass of ${}_{3}L{i}^7$

Where,

$M_p$ is the mass of the proton

$M_n$ is the mass of the neutron

$△m=$ $(3$ X $1.007825+4$ X $1.008898)-7=0.059067$

$△E=0.059067$ X $931=54.991377$

$E=\frac{△E}{A}=\frac{54.991377}{7}=7.736MeV$