Question 69: Atoms & Nuclei

Question

Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $\alpha$-decay with same $Q$ value of 1 MeV . The ratio of energies of $\alpha$-rays produced by A and B is $\_\_\_\_$ .

Accepted Answer

In $\alpha$-decay, a parent nucleus (X) of mass number A decays into a daughter nucleus (Y) and an $\alpha$-particle (Helium nucleus, mass number 4).
$$ \mathrm{X}_{\mathrm{Z}}^{\mathrm{A}} ightarrow \mathrm{Y}_{\mathrm{Z}-2}^{\mathrm{A}-4}+\mathrm{He}_2^4 $$
Using conservation of momentum, the momentum of the $\alpha$-particle ( $\mathrm{p}_\alpha$ ) must be equal and opposite to the momentum of the daughter nucleus $\left(\mathrm{p}_{\mathrm{D}}ight)$.
And using energy balance ( Q -value) the total kinetic energy released ( Q ) is shared between the $\alpha$-particle and the daughter nucleus.
$$ Q=K_\alpha+K_D $$
Kinetic energy can be expressed in terms of momentum (p) and mass (m) as $\mathrm{K}=\frac{\mathrm{p}^2}{2 \mathrm{~m}}$.
Since $\mathrm{p}_\alpha=\mathrm{p}_{\mathrm{D}}$, we have :
$\mathrm{K}_{\mathrm{D}}=\frac{\mathrm{p}^2}{2 \mathrm{~m}_{\mathrm{D}}}$ and $\mathrm{K}_\alpha=\frac{\mathrm{p}^2}{2 \mathrm{~m}_\alpha}$
Therefore, $\frac{\mathrm{K}_{\mathrm{D}}}{\mathrm{K}_\alpha}=\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}$.
Substituting $\mathrm{K}_{\mathrm{D}}=\mathrm{K}_\alpha\left(\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}ight)$ into the Q -value equation:
$$ \mathrm{Q}=\mathrm{K}_\alpha+\mathrm{K}_\alpha\left(\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}ight)=\mathrm{K}_\alpha\left(1+\frac{\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}ight)=\mathrm{K}_\alpha\left(\frac{\mathrm{m}_{\mathrm{D}}+\mathrm{m}_\alpha}{\mathrm{m}_{\mathrm{D}}}ight) $$
Since $\mathrm{m}_{\mathrm{D}}+\mathrm{m}_\alpha \approx \mathrm{A}$ (mass number of parent) and $\mathrm{m}_{\mathrm{D}} \approx \mathrm{A}-4$ :
$$ \mathrm{K}_\alpha=\mathrm{Q}\left(\frac{\mathrm{~A}-4}{\mathrm{~A}}ight) $$
The Q-value is the same ( 1 MeV ) for both substances.
For Substance A ( $\mathrm{A}_{\mathrm{A}}=200$ ) :
$$ K_{\alpha A}=Q\left(\frac{200-4}{200}ight)=Q\left(\frac{196}{200}ight) $$
For Substance $B\left(A_B=212ight)$ :
$$ K_{\alpha B}=Q\left(\frac{212-4}{212}ight)=Q\left(\frac{208}{212}ight) $$
So, the ratio is;
$$ \frac{K_{\alpha A}}{K_{\alpha B}}=\frac{Q\left(\frac{196}{200}ight)}{Q\left(\frac{208}{212}ight)}=\frac{196}{200} 	imes \frac{212}{208} $$
$$\Rightarrow $$ $$ \frac{\mathrm{K}_{\alpha \mathrm{A}}}{\mathrm{~K}_{\alpha \mathrm{B}}}=\frac{49}{50} 	imes \frac{53}{52}=\frac{49 	imes 53}{50 	imes 52}=\frac{2597}{2600} $$
Therefore, the ratio of energies of $\alpha$-rays produced by A and B is $\frac{2597}{2600}$.
Hence, the correct option is (C).

Answer

$\frac{2548}{2650}$

Answer

$\frac{2706}{2646}$

Answer

$\frac{2597}{2600}$

Answer

$\frac{2862}{2499}$

Atoms & Nuclei

Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous \alpha-decay with same Q value of 1 MeV . The ratio of energies of \alpha-rays produced by A and B is \_\_\_\_ .

Choose the correct answer: