(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?

 

 

 

 
 
 
 
 

Answers (1)

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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