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The alpha-decay of nucleus X²³⁰ produces two groups of alpha-particles: group 1 with kinetic energy 6.8 MeV (daughter in ground state) and group 2 with kinetic energy 5.2 MeV (daughter in excited state). Find the energy of the gamma photon emitted during de-excitation of the daughter nucleus.

1. 1.43 MeV
2. 1.53 MeV
3. 1.63 MeV
4. 1.73 MeV

Correct answer: 1.63 MeV

Solution

The excitation energy of the daughter = Q1 - Q2. Since Q = KE_alpha*(A/(A-4)) for alpha decay from nucleus of mass number A=230: Q1 = 6.8*(230/226) ≈ 6.92 MeV, Q2 = 5.2*(230/226) ≈ 5.29 MeV. E_gamma = Q1-Q2 = 1.6*(230/226) ≈ 1.63 MeV.

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