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  • An electron of mass ' m ', when accelerated through a potential V has de-Broglie wavelength

An electron of mass ' m ', when accelerated through a potential V has de-Broglie wavelength \lambda _e. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:           

Option: 1

\lambda_e \sqrt{\frac{M}{m}}

 

 

 


Option: 2

\lambda_e \sqrt{\frac{m}{M}}


Option: 3

\lambda_e \left ( \frac{M}{m} \right )


Option: 4

\lambda_e \left ( \frac{m}{M} \right )


Answers (1)

best_answer

As \lambda = \frac{h}{\sqrt{2mK}}

K=qV is same for both proton and electron

So  \frac{\lambda _{e}}{\lambda _{p}}= \sqrt{\frac{M}{m}}

So \lambda_p=\lambda_e \sqrt{\frac{m}{M}}

 

Posted by

Rishi

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