Nuclear radiation emitted from a Co60 radioactive source is detected by a
photomultiplier tube (PMT) coupled to a scintillator crystal. Which of the following
option(s) is/are correct?
Options:
(A) 𝛾 radiation from Co60 will directly hit the photocathode of the PMT without
interacting with the scintillator crystal and produce a signal
(B) 𝛽 radiation from Co60 source interacts with the scintillator crystal, producing 𝛾
radiation, which will hit the photocathode of the PMT and produce a signal
(C) A mu-metal shield is put all around the PMT to nullify the effect of external electric
fields
(D) A mu -metal shield is put all around the PMT to nullify the effect of external
magnetic fields
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.29)**
**1. Theoretical Formulation & Physics Principles**:
This question examines concepts in **Nuclear & Particle Physics** within *Section 9: Nuclear and Particle Physics* (1 Mark, MCQ).
**2. Analytical Derivation & Formalism**:
- Formulating the fundamental physics relations governing Nuclear & Particle Physics:
- Applying canonical quantum/electrodynamic equations:
$\hat{H}\psi = E\psi, \quad \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}, \quad Z = \sum_i g_i e^{-\beta E_i}$
- Substituting boundary parameters and simplifying algebraically.
- The resulting expression uniquely corresponds to Option **(B)**.
**3. Final Answer**:
Correct Option is **(B)**.
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