91Ó°ÊÓ

The de Broglie wavelength is given by,

$\mathit{λ}\mathbf{=}\frac{\mathbf{h}}{\mathbf{p}}$

Here, h is the Planck's constant, $\mathit{λ}$is wavelength, and p is the momentum.

The kinetic energy in terms of momentum is given by,

role="math" localid="1663142299185" $$\begin{gathered} \mathit{K}\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}} \\ \mathit{p}\mathbf{=}\sqrt{\mathbf{2}\mathbf{m}\mathbf{K}} \end{gathered}$$

Here, m is the mass.

Consider an electron, accelerated with a potential difference of V, the kinetic energy is given by,

$\mathit{K}\mathbf{}\mathbf{=}\mathbf{}\mathit{e}\mathit{V}$

The momentum is given by,

$\mathit{p}\mathbf{=}\sqrt{\mathbf{2}\mathbf{m}\mathbf{e}\mathbf{V}}$

The expression for wavelength will be as follows:

$\mathit{λ}\mathbf{=}\frac{\mathbf{h}}{\sqrt{\mathbf{2}\mathbf{m}\mathbf{e}\mathbf{V}}}$

From the above de Broglie wavelength expression, the wavelength is inversely proportional to the $\sqrt{V}$, which means the wavelength decreases as the potential difference increases and vice-versa.

As given the potential difference,

$$\begin{gathered} V_1=-100V \\ {V}_2=-200V \\ {V}_3=+100V \end{gathered}$$

So, the rank of de Broglie wavelength will be ${\mathrm{λ}}_2>{\mathrm{λ}}_1>{\mathrm{λ}}_3$.

Therefore, the rank is ${\mathrm{λ}}_2>{\mathrm{λ}}_1>{\mathrm{λ}}_3$.