In the nuclear decay given below:
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \rightarrow { }_{\mathrm{Z}+1}^{\mathrm{A}} \mathrm{Y}\rightarrow { }_{\mathrm{Z-1}}^{\mathrm{A-4}} \mathrm{B}\rightarrow { }_{\mathrm{Z-1}}^{\mathrm{A-4}} \mathrm{B}\) the particles emitted in the sequence are:

1.	\(\beta, \alpha, \gamma\)	2.	\( \gamma, \beta, \alpha\)
3.	\(\beta, \gamma,\alpha\)	4.	\(\alpha,\beta, \gamma\)

The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an:

The decay constants of two radioactive materials X₁ and X₂ are \(5\lambda\) and \(\lambda\) respectively. Initially, they have the same number of nuclei.  The ratio of the number of nuclei of X₁ to that of X₂  will be \(1/e\) after a time:
1. \(\lambda\)

2. \(\frac{1}{2\lambda }\)

3. \(\frac{1}{4\lambda }\)

4. \(\frac{e}{\lambda }\)

If \(M(A,~Z)\), \(M_p\)${}_{}$, and \(M_n\) denote the masses of the nucleus \(^{A}_{Z}X,\) proton, and neutron respectively in units of \(u\) \((1~u=931.5~\text{MeV/c}^2)\) and represent its binding energy \((BE)\) in \(\text{MeV}\). Then:

Two nuclei have their mass numbers in the ratio of \(1:3.\) The ratio of their nuclear densities would be:
1. \(1:3\)
2. \(3:1\)
3. \((3)^{1/3}:1\)
4. \(1:1\)

In the radioactive decay process, the negatively charged emitted β-particles are:

A nucleus ${}_Z{X}^A$ has a mass represented by \(M(A, Z).\) If \(M_P\) and \(M_n\) denote the mass of proton and neutron respectively and BE the binding energy, then:

1. $BE=\left[M\right(A,Z)-Z{M}_p-(A-Z\left)M_n\right]c^2$

2. $BE=[Z{M}_p+(A-Z)M_n-M(A,$ $Z\left)\right]c^2$

3. $BE=[Z{M}_p+A{M}_n-M-(A,$ $Z\left)\right]c^2$

4. $BE=M(A,$ $Z)-Z{M}_p-(A-Z)M_n$

The binding energy of deuteron is \(2.2~\text{MeV}\) and that of \(_2\mathrm{He}^{4}\) is \(28~\text{MeV}\). If two deuterons are fused to form one \(_{2}\mathrm{He}^{4}\), ${\mathrm{}}^{}$then the energy released is:
1. \(25.8~\text{MeV}\)
2. \(23.6~\text{MeV}\)
3. \(19.2~\text{MeV}\)
4. \(30.2~\text{MeV}\)

In a radioactive material, the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is λ, then:

1. $R_1=R_2{e}^{\lambda (t_1\mathit{+}{t}_2)}$

2. $R_1=R_2{e}^{-\lambda (t_1-t_2)}$

3. $R_1=R_2(t_1-t_2)$

4. $R_1=R_2$

The mass of a ${}_3{}^7\mathrm{Li}$ nucleus is \(0.042~\text{u}\) 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. \(4.6~\text{MeV}\)
2. \(5.6~\text{MeV}\)
3. \(3.9~\text{MeV}\)
4. \(23~\text{MeV}\)