(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.

 

 

 
 
 
 
 

Answers (1)
S safeer

(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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