If λ 1 and λ 2 denote the de-Broglie wavelengths of two particles with same masses but charges in the ratio of 1 : 2 after they are accelerated from rest through the same potential difference then

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TestHub · Accepted Answer

As, the expression for the de-Broglie wavelength associated with a charged particle having charge $Q$ and mass $m$ is

$λ = \frac{h}{\sqrt{2 m q V}}$

Where, $V =$ potential through which the particle is accelerated.

Now, given

$m_{1} = m_{2} = m$

and $\frac{q_{1}}{q_{2}} = \frac{1}{2}$

$\Rightarrow q_{2} = 2 q_{1}$

Then, $\frac{λ_{1}}{λ_{2}} = \frac{\frac{h}{\sqrt{2 m q V}}}{\frac{h}{\sqrt{4 m q V}}} = \frac{\sqrt{2}}{1}$

Hence, $λ_{1} = \sqrt{2} λ_{2}$

$\Rightarrow λ_{1} > λ_{2}$

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