# Plot a graph showing variation of de-Broglie wavelength $(\lambda )$ associated with a charged particle mass $m$, versus $\frac{1}{\sqrt{V}}$,Where $V$ is the potential difference through which the particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle

Let us consider a particle of charge $q$ and mass $m.$

To gain speed $v$ and kinetic energy $\inline KE,$ Let it be accelerated by a potential difference $\inline V$

Thus,

$KE =qV=\frac{1}{2}mv^2$

Now, on Rearranging the expression, we have

$v=\sqrt{\frac{2qV}{m}}$

The de-Broglie wavelength equation :-

$\lambda =\frac{h}{mv}$

On substituting the value of $v,$ we have

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

Now, we know the equation of straight line.-

$y=mx,$ Where $m$ is the shape of the line.

Hence, on plotting the graph of $\lambda$ vs $\frac{1}{\sqrt{V}}$.

We will have a straight line with a slope.

$S=\frac{h}{\sqrt{2mq}}$

On rearranging, we have

$q=\frac{h^{2}}{2mS^{2}}$

The graph of $\lambda$ vs $\frac{1}{\sqrt{V}}$. Can be drawn as :

Hence, we can determine $S$ from the graph, and this graph gives us the information regarding the magnitude of the charge of the particle.

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