Two radioactive substances A and B of mass numbers 200 and 212...

In an $\alpha$-decay, the kinetic energy of the emitted $\alpha$-particle is given by the conservation of momentum and energy as:

$K_{\alpha} = \dfrac{A - 4}{A} Q$

where $A$ is the mass number of the parent nucleus.

For substance A ($A = 200$):

$K_{\alpha, A} = \dfrac{200 - 4}{200} Q = \dfrac{196}{200} Q$

For substance B ($A = 212$):

$K_{\alpha, B} = \dfrac{212 - 4}{212} Q = \dfrac{208}{212} Q$

The ratio of the energies of the $\alpha$-rays produced by A and B is:

$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{\dfrac{196}{200} Q}{\dfrac{208}{212} Q} = \dfrac{196}{200} \times \dfrac{212}{208}$

Simplifying the fractions:

$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{49}{50} \times \dfrac{53}{52} = \dfrac{2597}{2600}$

Answer: $\dfrac{2597}{2600}$