# (a) An electron and a proton are accelerated through the same potential. Which one of the two has(i)     greater value of de-Broglie wavelength associated with it, and(ii)     lesser momentum? Justify your answer in each case.(b) How is the momentum of a particle related with its de-Broglie wavelength? Show the variation on a graph.

(a) We know the de-Broglie wavelength,

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

Since, $\lambda \alpha \frac{1}{\sqrt{m}}$     (For any other variable constant)

$\because m_{p}> m_{e}$

therefore,     $\lambda _{electron}> \lambda_{proton}$

(ii) We know , momentum

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

$\lambda _{electron}> \lambda _{proton}$

Hence, the momentum of the electron is less than the momentum of the proton.

(b) The relation between momentum and De-Broglie wavelength is given as;

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

Where

$p=$ momentum

$\lambda =$ wavelength

$h=$ planks constant.

Hence, the graph between $p$ and $\lambda$ is given as ;

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