# Potential energy of a particle varies as a function of $x$ and is given as $U(x)$ $\begin {bmatrix} E_{\circ} & 0 \leq x \leq 1 \\ 0 & x > 1 \end {bmatrix}$ The de-Broglie wavelengths of the particle in the region $0 \leq x \leq1$ and $x>1$ are $\lambda_1$ and $\lambda_2$ respectively. If the total energy of the particle is $2E_{\circ}$, the ratio $\large\frac{\lambda_1}{\lambda_2}$is

$\begin {array} {1 1} (a)\;\large\frac{1}{\sqrt 2} & \quad (b)\; \sqrt 2 \\ (c)\;2 & \quad (d)\;\large\frac{1}{2} \end {array}$

Ans : (b)
Total energy = Kinetic energy + Potential energy
For the region $0 \leq x \leq 1,$
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – E_{\circ} = E_{\circ}$
So, $\lambda_1 = \large\frac{h}{ \sqrt {2m(E_{\circ})}}$…………..(i)
For the region $x>1$,
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – 0 =2 E_{\circ}$
So, $\lambda_2 = \large\frac{h}{ \sqrt{2m(2E_{\circ})}}$…………….(ii)
From (i) & (ii),
$\large\frac{ \lambda_1}{ \lambda_2} = \sqrt 2$

edited Mar 13, 2014