The P.E. ' $U$ ' of a moving particle of mass ' $m$ ' varies with ' $\mathrm{x}$ ' is shown in the figure.…
The P.E. ' $U$ ' of a moving particle of mass ' $m$ ' varies with ' $\mathrm{x}$ ' is shown in the figure. The de-Broglie wavelength of the particle in the regions $0 \leq x \leq 1$ and $x>1$ are $\lambda_1$ and $\lambda_2$ respectively. If the total energy of the particle is ' $\mathrm{nE}$ ', then the ratio $\lambda_1 / \lambda_2$ is
$\sqrt{\frac{n^2}{n-1}}$
$\sqrt{\frac{n-1}{n}}$
$\sqrt{\frac{n}{n-1}}$
$\sqrt{\frac{n(n-1)}{n}}$
Solution
In the region $0 \leq x \leq 1$, the potential energy of the particle is $E$. Total energy is $\mathrm{nE}$.
Hence, kinetic energy, $\mathrm{K}=\mathrm{nE}-\mathrm{E}=(\mathrm{n}-1) \mathrm{E}$
Its momentum, $\mathrm{p}_1=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}$
$\therefore \lambda_1=\frac{\mathrm{h}}{\mathrm{p}_1}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{n}-1) \mathrm{E}}}$
In the region $\mathrm{x}>1, \mathrm{PE}$ is zero
Hence, total is kinetic energy.
$\begin{aligned}
& \therefore \mathrm{K}=\mathrm{nE} \\
& \therefore \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mnE}}} \\
& \therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}
\end{aligned}$
