The P.E. 'U' of a moving particle of mass 'm' varies with 'x'-axis as shown in figure. The deBroglie wavelength of the particle in the regions $0 \le x \le 1$ and $x > 1$ are $\lambda_1$ and $\lambda_2$ respectively. If the total energy of the particle is 'nE', then the ratio $\lambda_1 / \lambda_2$ is
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Always remember that the de Broglie wavelength depends purely on momentum (and thus kinetic energy), not total energy. Since $\lambda \propto 1/\sqrt{K}$, identifying the kinetic energy in each region immediately gives you the answer.
Step 1: Understand the question.
A particle moves along the $x$-axis. In one region its potential energy is one value, and in another region it is a different value. The total energy is $nE$. The de Broglie wavelength is $\lambda_1$ in the region $0 \le x \le 1$ and $\lambda_2$ in the region $x > 1$. We must find the ratio $\lambda_1/\lambda_2$. From the graph, the potential energy is $E$ in the first region and $0$ in the second region.
Step 2: Recall how total energy splits.
The total energy is the kinetic energy plus the potential energy: total $= K + U$. So the kinetic energy is total minus potential: $K = \text{total} - U$.
Step 3: Recall the wavelength rule.
The de Broglie wavelength is:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
So a bigger kinetic energy means a shorter wavelength. In short, $\lambda$ is one over the square root of $K$.
Step 4: Kinetic energy in region 1.
Here $U_1 = E$, so:
\[ K_1 = nE - E = E(n-1) \]
Thus $\lambda_1 = \dfrac{h}{\sqrt{2mE(n-1)}}$.
Step 5: Kinetic energy in region 2.
Here $U_2 = 0$, so:
\[ K_2 = nE - 0 = nE \]
Thus $\lambda_2 = \dfrac{h}{\sqrt{2mnE}}$.
Step 6: Take the ratio.
Divide the two wavelengths. The $h$ and $2m$ cancel:
\[ \frac{\lambda_1}{\lambda_2} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{nE}{E(n-1)}} = \sqrt{\frac{n}{n-1}} \]
This matches option (1).
\[ \boxed{\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{n}{n-1}}} \]
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