A nucleus represented by the symbol ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$ has:

Which of the following pairs of nuclei are isotones?

1. ${}_{34}{}^{74}\mathrm{Se},$ ${}_{31}{}^{71}\mathrm{Ca}$          

2. ${}_{42}{}^{92}\mathrm{Mo},$ ${}_{40}{}^{92}\mathrm{Zr}$

3. ${}_{38}{}^{81}\mathrm{Sr},$ ${}_{38}{}^{86}\mathrm{Sr}$             

4. ${}_{20}{}^{40}\mathrm{Ca},$ ${}_{16}{}^{32}\mathrm{S}$

The binding energy of deuteron is 2.2 MeV and that of ${}_{2}H{e}^4$ is 28 MeV. If two deuterons are fused to form one ${}_{2}H{e}^4$ then the energy released is:

The binding energies of the nuclei A and B are E_a and E_b respectively. If three atoms of the element B fuse to give one atom of element A and an energy Q is released, then E_a, E_b and Q are related as:

1. E_a – 3E_b = Q

2. 3E_b – E_a = Q

3. E_a + 3E_b = Q

4. E_b + 3E_a = Q

If M (A, Z), ${\mathrm{M}}_p$, and ${\mathrm{M}}_n$ denote the masses of the nucleus ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$, proton, and neutron respectively in units of u (1 u = 931.5 MeV/c²) and BE represents its binding energy in MeV, then:

The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form a helium atom, the energy released is:

1. 19.2 MeV

2. 23.6 MeV

3. 26.9 MeV 

4. 13.9 MeV

In the reaction

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

1.  c + a - b

2.  c - a - b

3.  a + b + c

4.  a + b - c

The mass of a ${}_3{}^7\mathrm{Li}$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of the ${}_3{}^7\mathrm{Li}$ nucleus is near:

1. 4.6 MeV

2. 5.6 MeV

3. 3.9 MeV

4. 23 MeV

The Binding energy per nucleon of \(^{7}_{3}\mathrm{Li}\) and \(^{4}_{2}\mathrm{He}\) nucleon are \(5.60~\text{MeV}\) and \(7.06~\text{MeV}\), respectively. In the nuclear reaction \(^{7}_{3}\mathrm{Li} + ^{1}_{1}\mathrm{H} \rightarrow ^{4}_{2}\mathrm{He} + ^{4}_{2}\mathrm{He} +Q\), the value of energy \(Q\) released is:
1. \(19.6~\text{MeV}\)
 2. \(-2.4~\text{MeV}\)
 3. \(8.4~\text{MeV}\)
4. \(17.3~\text{MeV}\)