Beta Decay Involves Three-Body Emission

Steps:

• In β⁻ decay, carbon-14 transforms into nitrogen-14 by emitting an electron (β particle) and an antineutrino.
• The total energy released (Q-value) is fixed for all carbon-14 nuclei, based on mass difference.
• This energy is shared among the recoiling nitrogen-14 nucleus, the β particle, and the antineutrino.
• Since the antineutrino carries variable kinetic energy, the β particle's kinetic energy varies accordingly, producing a continuous spectrum.

Why D is correct:

• In beta decay, the three-body process (nucleus + β + ν-bar) follows conservation of energy and momentum, distributing the fixed Q-value variably among particles, as per Fermi's theory.

Why the others are wrong:

• A: All carbon-14 nuclei are identical isotopes with the same mass.
• B: β particles are electrons with fixed, identical masses.
• C: The Q-value is constant for the specific nuclear transition.

Final answer: D