The mass defect of $ { }_{2}^{4} He $ is $ 0.03 \ u $. The binding energy per nucleon of helium (in $ MeV $ ) is

$1 \text{ amu}$ is equal to

Binding energy per nucleon of ${ }_1^2 H$ and ${ }_2^4 He$ are $1.1 \ MeV$ and $7.0 \ MeV$ respectively. Energy released in the process ${ }_1^2 H + { }_1^2 H \rightarrow { }_2^4 He$ is: (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 mass defect for the nucleus of helium is $0.0303 \, a.m.u.$ What is the binding energy per nucleon for helium in $MeV$?

The $Q$-value of a nuclear reaction and kinetic energy of the projectile particle,$K_{p}$,are related as