GATE 2022 PH – Question 49

Correct answer: (A) There are two equilibrium points along the radial coordinate; (B) There is one stable equilibrium point at rଵ and one unstable equilibrium point at rଶ> rଵ

Explanation

For a unit mass with $V(r)=-V_0e^{-r^2}$ and angular momentum $L$, the radial motion occurs in the **effective potential**
$$V_{eff}(r)=\frac{L^2}{2r^2}-V_0e^{-r^2}.$$

With $L=0.5\sqrt{V_0}$: $\dfrac{L^2}{2}=\dfrac{V_0}{8}$, so
$$V_{eff}=V_0\left[\frac1{8r^2}-e^{-r^2}\right].$$

**Equilibrium points:** $V_{eff}^\prime=0$:
$$-\frac{1}{4r^3}+2re^{-r^2}=0\;\Rightarrow\;2r^4e^{-r^2}=\frac14 .$$

The function $f(r)=2r^4e^{-r^2}$ starts from 0, rises to a maximum at $r^2=2$ (where $f=8e^{-2}\approx1.08>\tfrac14$) and then falls back towards 0. It therefore crosses the value $\tfrac14$ at **two** radii, $r_1<r_2$.

**Stability.** Near $r\to0$ the centrifugal term makes $V_{eff}$ large and positive (decreasing with $r$), then the potential well takes over. The sign of $V_{eff}^\prime$ goes from negative to positive at $r_1$ (a **minimum**, stable) and from positive to negative at $r_2$ (a **maximum**, unstable).

So there are two equilibrium points, one stable at $r_1$ and one unstable at $r_2>r_1$. Answer **A and B**.