A particle of unit mass moves in a potential V(r) = -V_0e-^2
. If the angular
momentum of the particle is L = 0.5ඥV_0 , then which of the following statements
are true?
Options:
(A) There are two equilibrium points along the radial coordinate
(B) There is one stable equilibrium point at r1 and one unstable equilibrium point at
r2 > r1
(C) There are two stable equilibrium points along the radial coordinate
(D) There is only one equilibrium point along the radial coordinate
Official Answer Key:
A,B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.49)**
**1. Classical Mechanics (Central Force Effective Potential)**:
- Mass $m = 1$, angular momentum $L = 0.5\sqrt{V_0}$, potential $V(r) = -V_0 e^{-r^2}$.
- Effective potential:
$V_{\text{eff}}(r) = \frac{L^2}{2m r^2} + V(r) = \frac{(0.5\sqrt{V_0})^2}{2(1)r^2} - V_0 e^{-r^2} = \frac{V_0}{8r^2} - V_0 e^{-r^2}$
**2. Detailed Step-by-Step Equilibrium Analysis**:
- Setting first derivative to zero:
$\frac{dV_{\text{eff}}}{dr} = -\frac{V_0}{4r^3} + 2V_0 r e^{-r^2} = 0$
$\implies 2r e^{-r^2} = \frac{1}{4r^3} \implies r^4 e^{-r^2} = \frac{1}{8} = 0.125$
- Let $f(r) = r^4 e^{-r^2}$:
- $f(0) = 0$
- At $r = \sqrt{2}$, $f(\sqrt{2}) = 4 e^{-2} = \frac{4}{7.389} \approx 0.541$
- As $r \to \infty$, $f(r) \to 0$.
- Since the maximum value ($0.541$) exceeds $0.125$, the horizontal line $0.125$ intersects $f(r)$ at **two distinct radial equilibrium points**:
- $r_1 < \sqrt{2}$ where slope is positive (local minimum of $V_{\text{eff}} \implies$ **stable equilibrium**).
- $r_2 > \sqrt{2}$ where slope is negative (local maximum of $V_{\text{eff}} \implies$ **unstable equilibrium**).
- Therefore, there are two equilibrium points, one stable at $r_1$ and one unstable at $r_2 > r_1$.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (A, B)**.
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