# (a) Plot a graph showing the variation of de Broglie wavelength $(\lambda )$ associated with a  charged particle of mass m, versus $\sqrt{V}$ , where  V is the accelerating potential.(b)   An electron, a proton and an alpha particle have the same kinetic energy. Which one has the shortest wavelength?

The relation between De Broglie wavelength and potential is given by

$\lambda =\frac{h}{\sqrt{2mqV}}$

$\lambda\propto\frac{1}{\sqrt{V}}$

So the graph is as follows

(b)

$\lambda =\frac{h}{\sqrt{2mqV}}$

Therefore, now since the electron has the lowest mass, so, $\lambda$ for an electron is maximum followed by proton then alpha particle.

$\lambda\; \; electron> \lambda\; \; proton > \lambda _{\alpha }-particle$

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