The mass of a proton is $1.0073 \; u$ and that of a neutron is $1.0087 \; u$ ($u =$ atomic mass unit). The binding energy of ${ }_2^4 \text{He}$ is (Given: helium nucleus mass $\approx 4.0015 \; u$):

The binding energy per nucleon of ${O^{16}}$ is $7.97 \,MeV$ and that of ${O^{17}}$ is $7.75 \,MeV$. The energy (in $MeV$) required to remove a neutron from ${O^{17}}$ is

The energy released in a nuclear reaction is due to:

Derive the formula for the binding energy of a nucleus by explaining the mass defect of the nucleus and write the formula for the binding energy per nucleon.

Consider the following nuclear reaction : $^{238}\text{U} \rightarrow ^{234}\text{Th} + ^4\text{He}$. Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^4\text{He}$ as $238.050 \text{ u}$, $234.043 \text{ u}$ and $4.003 \text{ u}$, respectively. The $Q$ value for the reaction, in $\text{keV}$, is : [Given : $1 \text{ u} = 931.5 \text{ MeV/c}^2$]

In the nuclear reaction $_1^2H + _1^3H \to _2^4He + _0^1n$, if the binding energies of $_1^2H$, $_1^3H$, and $_2^4He$ are respectively $a$, $b$, and $c$ (in $MeV$), then the energy (in $MeV$) released in this reaction is: