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Question:

The isotope ¹²⁵B having a mass 12.014 u undergoes β⁻-decay to ¹²⁶C. ¹²⁶C has an excited state of the nucleus (¹²⁵C*) at 4.041 MeV above its ground state. If ¹²⁵B decays to ¹²⁶C*, the maximum kinetic energy of the β⁻-particle in units of MeV is (1 u = 931.5 MeV/c², where c is the speed of light in vacuum).

Solution:

¹²⁵B → ¹²⁶C + β⁻ + ḡ
Q = (m₁₂₅B - m₁₂₆C)c² = (m₁₂₅B - (m₁₂₆C + Δm))c² = (m₁₂₅B - m₁₂₆C)c² - Δmc² = 0.014 × 931.5 - 4.041 = 9 MeV