Question

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

Answer 1

For ${}_3{Li}^7$ nucleus

Mass defect, $△M=0.0424$

$\because 1u=931.5MeV/c^2$

$\therefore △M=0.042\times 931.5MeV/c^2$

$=39.1MeV/c^2$

Binding energy, $E_b=△M{c}^2$

$=(39.1\times \frac{MeV}{c^2})c^2$

$=39.1MeV$