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:

If in a nuclear fusion process. the masses of the fusing nuclei be \(m_1\) and \(m_2\) and the mass of the resultant nucleus be \(m_3,\) then:

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 ${}_2{}^4\mathrm{He}$ is: (Given: helium nucleus mass ≈ 4.0015 u)

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 a fission reaction,

$$\(^{236}_{92}U\rightarrow ~^{117}X~+~^{117}Y~+~^1_0n~+~^1_0n,\) the binding energy per nucleon of X and Y is 8.5 MeV whereas that of \(^{236}U\) is 7.6 MeV. The total energy liberated will be about:

1. 2000 MeV

2. 200 MeV

3. 2 MeV 

4. 1 keV

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