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

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.
\(
\begin{aligned}
& \therefore \mathrm{K}=\mathrm{nE} \\
& \therefore \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mnE}}} \\
& \therefore \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{n}}{\mathrm{n}-1}}
\end{aligned}
\)