Question

A particle of unit mass is released from origin with a velocity $\mathbf{v} = v_0 \hat{i}$ inside a potential well. The potential energy associated with this well is of the form $\phi(x) = ax^2\exp(-bx^2)$ where $a, b$ are equal to unity in SI units. The minimum value of $v_0$, in SI units, for the particle to cross the potential well is closest to

• A. 0.85
• B. 0.75
• C. 0.95
• D. 0.65

Answer 1

Answer: (A)

0.85

Answer 2

The energy conservation gives:

$$\frac{1}{2}v_0^2 = \phi_{\text{max}}$$

Given $\phi(x) = x^2 e^{-x^2}$ (since $a=b=1$). Find $x$ where $\phi(x)$ is maximum:

$$\frac{d\phi}{dx} = 2x e^{-x^2}(1-x^2)=0 \quad \Rightarrow \quad x=0 \text{ or } x=\pm1.$$

Discarding $x=0$ (minimum), maximum is at $x = \pm 1$.

Maximum potential:

$$\phi_{\text{max}}=\phi(1)=1^2 e^{-1}=\frac{1}{e}.$$

Thus,

$$\frac{1}{2}v_0^2 = \frac{1}{e} \quad \Rightarrow \quad v_0=\sqrt{\frac{2}{e}}.$$

Numerically,

$$v_0\approx\sqrt{\frac{2}{2.7183}}\approx\sqrt{0.7358}\approx0.857.$$