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. \( 1: 2^{1 / 3}\)

Activity of a radioactive sample decreases to \(\frac{1}{3}\)^rd of its original value in \(3\) days. Then, in \(9\) days its activity will become:

When Lithium ${(}^{7}Li)$ is bombarded by a proton, two alpha particles ${(}^{4}He)$ are produced. The masses of ${}^{7}Li,{ }^{1}H and{ }^{4}He$ are 7.016004 u, 1.007825 u and 4.002603 u respectively. The reaction energy is nearly:

1. 17 eV

2. 17 keV

3. 17 MeV

4. 170 MeV

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