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 ________.

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: \(\dfrac{2597}{2600}\)