QB_PH_SEC9_026: Nuclear Models & Decays

A free neutron undergoes beta decay via $n \to p + e^- + \bar{\nu}_e$ with a $Q$-value of $Q = 0.782\text{ MeV}$. Assuming the proton recoils with negligible kinetic energy, what is the maximum kinetic energy $T_{e,\text{max}}$ (in MeV) of the emitted electron? (Give your answer to three decimal places)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Kinematics of Three-Body Beta Decay**: $Q = T_p + T_e + T_\nu$ 2. **Endpoint Condition**: The electron acquires its maximum possible kinetic energy when the accompanying massless (or ultra-light) antineutrino carries zero momentum/energy ($T_\nu \to 0$): $Q = T_p + T_{e,\text{max}}$ 3. **Proton Recoil Consideration**: Because $m_p \approx 1836\, m_e$, maximum recoil kinetic energy of the proton is: $T_{p,\text{max}} \approx \frac{Q^2}{2 m_p c^2} = \frac{(0.782\text{ MeV})^2}{2(938.3\text{ MeV})} = 0.00032\text{ MeV} \approx 0.3\text{ keV}$ 4. **Evaluation**: To 3 decimal places, $T_{e,\text{max}} \approx Q = 0.782\text{ MeV}$. 5. **Official Range**: $[0.781, 0.783]$.