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:

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

Two radioactive substances A and B have decay constants 5λ and λ respectively. At t = 0, they have the same number of nuclei. The ratio of the number of nuclei of A to those of B will be $\frac{1}{e^2}$ after a time interval:

1. $\frac{1}{4\lambda }$

2. $4\lambda$

3. $2\lambda$

4. $\frac{1}{2\lambda }$

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. $R_1=R_2$

If M₀ is the mass of an oxygen isotope ₈O¹⁷, M_P and M_n are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is:

1.  ${\mathrm{M}}_{\mathrm{o}}{\mathrm{c}}^2$

2.  $\left({\mathrm{M}}_{\mathrm{o}}-17{\mathrm{M}}_{\mathrm{c}}\right) {\mathrm{c}}^2$

3.  $\left({\mathrm{M}}_{\mathrm{o}}-8{\mathrm{M}}_{\mathrm{p}}\right) {\mathrm{c}}^2$

4.  $\left(8{\mathrm{M}}_{\mathrm{p}}+9{\mathrm{M}}_{\mathrm{n}}-{\mathrm{M}}_{\mathrm{o}}\right) {\mathrm{c}}^2$

A nucleus disintegrates into two nuclear parts which have their velocities in the ratio \(2:1\). The ratio of their nuclear size will be:
1. \( 2^{1 / 3}: 1 \)
2. \( 1: 3^{1 / 2} \)
3. \( 3^{1 / 2}: 1 \)
4. \( 1: 2^{1 / 3}\)