JEE Mains · Physics · STD 12 - 13. 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 ________.
- A \(\dfrac{2548}{2650}\)
- B \(\dfrac{2706}{2646}\)
- C \(\dfrac{2597}{2600}\)
- D \(\dfrac{2862}{2499}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{2597}{2600}\)
Step-by-step Solution
Detailed explanation
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 (\(A = 200\)):…
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