GATE PH 2022 • Q48 MSQ • 2 Marks

Section 4: Quantum Mechanics

Schrodinger equation & Wave Mechanics

Subtopic: Wavefunction, Probability Density & Uncertainty Principle

Question Statement:

A particle of mass m is moving inside a hollow spherical shell of radius a so that the potential is V(r) = 0 for r < a ∞ for r ≥ a The ground state energy and wavefunction of the particle are E_0 and R(r), respectively. Then which of the following options are correct?

Options:

  • (A) $E_0 = \frac{\hbar^2 \pi^2}{2m a^2}$
  • (B) $-\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dR}{dr}\right) = E_0 R \quad (r < a)$
  • (C) $-\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{d^2 R}{dr^2} = E_0 R \quad (r < a)$
  • (D) $R(r) = \frac{1}{r} \sin\left(\frac{\pi r}{a}\right) \quad (r < a)$

Official Answer Key:

A,B,D

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.48)** **1. Quantum Mechanics Formulation (Spherical Shell Well)**: - Potential: $V(r) = 0$ for $r < a$ and $V(r) = \infty$ for $r \ge a$. - Stationary Schrödinger equation in spherical coordinates: $-\frac{\hbar^2}{2m} \nabla^2 \psi = E\psi$ - Radial equation for $u(r) = r R(r)$ with $l = 0$ (s-wave ground state): $-\frac{\hbar^2}{2m} \frac{1}{r^2} \frac{d}{dr}\left(r^2 \frac{dR}{dr}\right) = E_0 R(r) \quad (r < a)$ **Statement (B) is TRUE**. (Statement (C) lacks the proper metric factors). **2. Ground State Energy and Wavefunction**: - For $l = 0$, $u''(r) + k^2 u(r) = 0 \implies u(r) = A \sin(kr) + B \cos(kr)$. - Since $R(r) = u(r)/r$ must remain finite at $r = 0$, $B = 0 \implies u(r) = A\sin(kr)$. - Boundary condition at $r = a$: $u(a) = 0 \implies ka = \pi \implies k = \frac{\pi}{a}$. - Ground state radial wavefunction: $R(r) = \frac{1}{r} \sin\left(\frac{\pi r}{a}\right)$ **Statement (D) is TRUE**. - Ground state energy: $E_0 = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2 \pi^2}{2m a^2}$ **Statement (A) is TRUE**. **3. Official Verification**: The verified official IIT Master Answer is **Options (A, B, D)**.

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