A nucleus undergoes β⁻-decay. The neutral atomic mass of the parent exceeds the neutral atomic mass of the daughter by 0.0030 u. Neglect daughter recoil. Take 1 u·c² = 931.5 MeV. What is the maximum possible kinetic energy of the emitted electron?
Options

Question

A nucleus undergoes β⁻-decay. The neutral atomic mass of the parent exceeds the neutral atomic mass of the daughter by 0.0030 u. Neglect daughter recoil. Take 1 u·c² = 931.5 MeV. What is the maximum possible kinetic energy of the emitted electron?

Options

G1 Physics · Accepted Answer

For β⁻-decay, when neutral atomic masses are used, the Q-value is directly Q = [M(parent)−M(daughter)]c². Thus, Q = 0.0030×931.5 = 2.7945 MeV. The β-spectrum is continuous because energy is shared with the antineutrino and daughter recoil. The maximum electron kinetic energy occurs when the antineutrino carries negligibly small energy and daughter recoil is neglected. Therefore, K(max) ≈ 2.79 MeV. Correct option: C.