Questions: Beta Decay and Energy Conservation in Weak Interactions

Question 1

Alpha particles emitted from a given isotope always have a single discrete energy, but beta particles from the same decay mode have a continuous range of energies. What is the fundamental reason for this difference?

• A. Alpha particles are heavier and therefore lose energy more predictably as they travel through matter
• B. Beta decay involves a three-body final state (daughter nucleus, electron, and antineutrino), so the Q-value is shared among three particles in variable proportions
• C. Beta decay is governed by the weak force, which operates over a broader energy range than the strong force behind alpha decay
• D. The electron's small mass allows it to receive varying fractions of the Q-value due to quantum uncertainty

Answer: (B)

The number of final-state particles determines whether the energy spectrum is discrete or continuous. Alpha decay has two final-state particles (alpha particle + daughter nucleus): conservation of energy and momentum uniquely fixes each particle's energy for a given Q-value, producing a discrete line. Beta decay has three final-state particles: the Q-value is shared between the electron and the antineutrino in continuously variable proportions, with the daughter nucleus taking a small recoil. This three-body kinematics produces a continuous spectrum. The weak force (option C) and quantum uncertainty (option D) are not the reason — it is strictly the counting of final-state particles.

Question 2

A nucleus undergoes beta-plus decay (p → n + e⁺ + νe). What minimum Q-value is required for this decay to proceed?

• A. Q > 0 — any positive Q-value is sufficient, just as in beta-minus decay
• B. Q > mec² ≈ 0.511 MeV — one positron mass must be created
• C. Q > 2mec² ≈ 1.022 MeV — a positron and electron are created as a pair
• D. Q > mpc² — a proton mass must be converted to a neutron mass

Answer: (C)

Beta-plus decay creates a positron from rest mass-energy. Since the positron is an antiparticle not present in the initial nucleus, its rest mass (mec² ≈ 0.511 MeV) must come from the Q-value. Unlike beta-minus decay (where the electron was already present in the atomic electron cloud and not created from nuclear mass), beta-plus decay requires creating new mass. The minimum Q-value is 2mec² ≈ 1.022 MeV — the factor of 2 arises when working with atomic (rather than nuclear) masses, as the atomic mass accounting effectively requires paying for two electron masses. When Q < 1.022 MeV, electron capture (the competing process) occurs instead.

Question 3

In beta-minus decay, the maximum kinetic energy of the emitted electron is approximately equal to the Q-value of the decay.

• T. True
• F. False

Answer: True

True. The electron achieves maximum kinetic energy when the antineutrino carries away near-zero energy (and the daughter nucleus takes a small but negligible recoil). In the limit of a nearly massless neutrino carrying zero energy, essentially all the Q-value goes to the electron. This maximum electron energy — called the endpoint energy — is what is measured in beta spectroscopy and was the key experimental observable that revealed the continuous spectrum and the 'missing' energy in the first place. Pauli's neutrino hypothesis explained why most decays produce electrons with less than the maximum energy.

Question 4

In electron capture decay, the nucleus emits an inner-shell electron, which can be detected by particle detectors.

• T. True
• F. False

Answer: False

False. In electron capture, the nucleus *absorbs* an inner-shell atomic electron (p + e⁻ → n + νe) and emits a neutrino — no electron is emitted from the nucleus. The name describes what happens to the electron (it is captured), not that an electron is produced. After the inner-shell electron is absorbed, a higher-shell electron drops to fill the vacancy, emitting a characteristic X-ray — which *is* detectable, but is not an electron. This is a common source of confusion between electron capture and beta-minus decay.

Question 5

Why did the continuous energy spectrum of beta decay electrons appear to threaten the law of conservation of energy, and how did Pauli's neutrino hypothesis resolve the apparent violation?

Model answer: In a two-body decay (like alpha decay), the Q-value uniquely fixes the energy of each product via conservation of energy and momentum, producing a discrete spectral line. In beta decay, physicists initially assumed a two-body decay (electron + daughter nucleus). The observed continuous spectrum meant that different electrons emerged with different energies, with the 'missing' energy varying decay by decay — a seemingly random violation of conservation of energy. Pauli hypothesized a third particle (the neutrino) that is emitted simultaneously with the electron, with the Q-value split between them. The electron's energy is continuous because the neutrino's energy varies decay by decay, but in every single decay the total energy (electron + neutrino + nuclear recoil) equals the Q-value. Conservation of energy is preserved in every individual event; the neutrino's energy is just undetected.

Pauli's proposal was bold because the neutrino had never been observed and seemed to violate the spirit of Occam's razor. But the alternative — abandoning conservation of energy — was unacceptable to most physicists (though Niels Bohr temporarily proposed it). Fermi's 1934 theory quantitatively predicted the shape of the beta spectrum; the neutrino was directly detected by Cowan and Reines in 1956, confirming Pauli's hypothesis.