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Neon-23 undergoes beta-minus decay: 10Ne²³ -> 11Na²³ + beta⁻ + antineutrino. The atomic mass of Ne-23 is 22.9945 u and that of Na-23 is 22.9898 u. Given that 1 u = 931.5 MeV/c² and mass of electron = 0.511 MeV/c², find the maximum kinetic energy of the beta particle (in MeV), to the nearest integer.

1. 1
2. 2
3. 4
4. 5

Correct answer: 4

Solution

For beta-minus decay using atomic masses: Q = [M(Ne-23) - M(Na-23)] * 931.5 MeV/u. The electron masses cancel because the atomic mass includes electron masses. Mass difference = 22.9945 - 22.9898 = 0.0047 u. Q = 0.0047 * 931.5 = 4.378 MeV ~ 4 MeV. The maximum kinetic energy of the beta particle equals Q (when antineutrino gets zero energy).

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