Ratio of Alpha Particle Energies for Radioactive Decay

When a stationary radioactive parent nucleus with mass number $A$ undergoes $\alpha$-decay, the reaction is represented as: $X_A \longrightarrow Y_{A-4} + \alpha_4$

By the conservation of linear momentum, the magnitude of momentum of the emitted $\alpha$-particle ($p_\alpha$) and the recoil daughter nucleus ($p_Y$) are equal: $p_\alpha = p_Y$

The total disintegrating energy ($Q$-value) is shared between the kinetic energy of the $\alpha$-particle ($E_\alpha$) and the kinetic energy of the recoil daughter nucleus ($E_Y$): $Q = E_\alpha + E_Y$

Expressing kinetic energy in terms of momentum: $E_Y = \frac{p_Y^2}{2 m_Y} = \frac{p_\alpha^2}{2 m_Y} = E_\alpha \left(\frac{m_\alpha}{m_Y}\right)$

Substituting $E_Y$ into the $Q$-value equation: $Q = E_\alpha \left(1 + \frac{m_\alpha}{m_Y}\right)$

Approximating masses using mass numbers ($m_\alpha \approx 4$ and $m_Y \approx A - 4$): $Q = E_\alpha \left(1 + \frac{4}{A-4}\right) = E_\alpha \left(\frac{A}{A-4}\right)$

Thus, the energy of the emitted $\alpha$-particle is given by: $E_\alpha = Q \left(\frac{A-4}{A}\right)$

For substance A ($A_A = 200$): $E_{\alpha, A} = Q \left(\frac{200 - 4}{200}\right) = Q \left(\frac{196}{200}\right)$

For substance B ($A_B = 212$): $E_{\alpha, B} = Q \left(\frac{212 - 4}{212}\right) = Q \left(\frac{208}{212}\right)$

Since both processes have the same $Q$-value, the ratio of the energies of the $\alpha$-rays produced by A and B is: $\frac{E_{\alpha, A}}{E_{\alpha, B}} = \frac{\frac{196}{200}}{\frac{208}{212}} = \frac{196 \times 212}{200 \times 208}$

Simplifying the fractions: $\frac{196}{200} = \frac{49}{50}$ $\frac{212}{208} = \frac{53}{52}$

Therefore: $\frac{E_{\alpha, A}}{E_{\alpha, B}} = \frac{49 \times 53}{50 \times 52} = \frac{2597}{2600}$

This matches option C.