Question

Nuclear binding energy is the energy released during the hypothetical formation of the nucleus by the condensation of individual nucleons. Thus, binding energy per nucleon =$\frac{\text{Total binding energy}}{\text{Number of nucleons}}$ For example, the mass of hydrogen atom is equal to the sum of the masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons, neutrons and electrons present. This difference in mass, termed as mass defect, is a measure of the binding energy of protons and neutrons in the nucleus. The mass-energy relationship postulated by Einstein is expressed as: $\Delta E=\Delta m{c}^2$
Where $\Delta E$ is the energy liberated, $\Delta m$ is the loss of mass, and c is the speed of light.
In the reaction ${}_1^{2}H{+}_1^{3}H{\to }_2^{4}He{+}_0^{1}n$, if binding energies ${}_1^{2}H{,}_1^{3}H\text{and}{}_2^{4}He$ are respectively a, b and c (in MeV), then the energy released in this reaction is:

• A. $a+b+c$
• B. $a+b-c$
• C. $c-a-b$
• D. $c+a-b$

Answer 1

The correct option is C $c-a-b$

${}_1^{2}H{+}_1^{3}H\to {}_2^{4}He+{}_0^{1}n$
B.E $a$ $b$ $c$

mass defect = $\frac{B.E}{c^2}$
mass defect = $\frac{c-a-b}{c^2}$

$\Delta E=\Delta m\times c^2$
= $\frac{c-a-b}{c^2}\times c^2$
= $c-a-b$