If the binding energy of $N^{14}$ is $7.5 \text{ MeV}$ per nucleon and that of $N^{15}$ is $7.7 \text{ MeV}$ per nucleon, then the energy required to remove a neutron from $N^{15}$ is (in $\text{ MeV}$)

The mass defect for the nucleus of helium is $0.0303 \, a.m.u.$ What is the binding energy per nucleon for helium in $MeV$?

${ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q$
In the given nuclear reaction,the approximate amount of energy released will be $.....\,MeV$.
[Given: mass of ${ }_{92}^{238} A = 238.05079 \, u$,mass of ${ }_{90}^{234} B = 234.04363 \, u$,mass of ${ }_2^4 D = 4.00260 \, u$,and $1 \, u = 931.5 \, MeV/c^2$]

The binding energy per nucleon for ${O^{16}}$ and ${O^{17}}$ are $7.97 \, MeV$ and $7.75 \, MeV$ respectively. How much energy in $MeV$ is required to remove one neutron from ${O^{17}}$?

If the binding energy per nucleon of a nuclide is high,then:

The energy equivalent to a substance of mass $1 \,g$ is