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PhysicsInteger+3Question 15 of 54

JEE Advanced 2016 · Paper 1, Physics question 15

The isotope 512B^{12}_{5}\mathrm B having a mass 12.014 u undergoes β-decay to 612C^{12}_{6}\mathrm C. 612C^{12}_{6}\mathrm C has an excited state of the nucleus (612C∗)(^{12}_{6}\mathrm C^*) at 4.041 MeV above its ground state. If 512B^{12}_{5}\mathrm B decays to 612C∗^{12}_{6}\mathrm C^*, the maximum kinetic energy of the β-particle in units of MeV is (1 u=931.5 MeV/c21\ \mathrm u=931.5\ \mathrm{MeV}/c^2, where cc is the speed of light in vacuum).

Answer and worked solution

Official answer: 8

Q-value. For β⁻ decay, atomic masses give the energy release directly:
Q=(12.014−12.000)×931.5=13.04 MeV.Q=(12.014-12.000)\times931.5=13.04\text{ MeV}.

Decay to the excited state. The excited level takes 4.041 MeV:
Kmax=13.04−4.04≈9 MeV.K_{max}=13.04-4.04\approx9\text{ MeV}.
The official key accepts 8 or 9.

Answer: 9 (key: 8 or 9)

Topics:Mass defect and binding energyRadioactive decay
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