Subtopic: Relativistic Momentum, Energy & 4-Vectors
Question Statement:
Which of the following consideration (s) is/are showing t hat nuclear beta decay ,
n → p + e- + 𝜈̅e, has to be a three-body decay?
Options:
(A) Continuous distribution of the electron energy
(B) Spin of the final state
(C) Mass of the electron
(D) Mass of the proton
Official Answer Key:
A;B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.27)**
**1. Nuclear Beta Decay Kinematics & Conservation Laws**:
- Beta decay process:
$n \to p + e^- + \bar{\nu}_e$
**2. Physical Analysis of 3-Body Kinematics (MSQ)**:
- **(A) Continuous distribution of electron energy [TRUE]**:
- In a two-body decay $A \to B + C$, conservation of energy and linear momentum uniquely fixes the kinetic energy of both outgoing particles:
$E_C = \frac{m_A^2 + m_C^2 - m_B^2}{2m_A}c^2 = \text{discrete constant}$
- Historically, beta electrons were observed to possess a continuous energy spectrum from $0$ up to $E_{\max}$. This proved kinematically that a third invisible particle (the electron antineutrino $\bar{\nu}_e$) must share the available Q-value.
- **(B) Spin of the final state [TRUE]**:
- All three particles (neutron, proton, electron) are spin-1/2 fermions ($s = 1/2$).
- If decay were a two-body process ($n \to p + e^-$), the coupled final state angular momentum would be:
$\frac{1}{2} \otimes \frac{1}{2} = 0 \oplus 1 \quad (\text{strictly integer spin})$
- Since the initial neutron has half-integer spin $1/2$, angular momentum conservation would be violated. A third spin-1/2 particle ($ar{\nu}_e$) is required to form a half-integer total final spin.
- **(C) & (D) False**: Electron and proton rest masses do not dictate multi-body kinematics.
**3. Official Verification**:
The verified official IIT answer is **Options (A) and (B)** (`['A', 'B']`).
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.