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An electron and proton are moving in the same direction and possess same K.E. Find the ratio of de-Broglie wavelength associated with these particles. |
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Answer» Solution :For photon `K.E., E_(p) =1/2m_(p).v_(p)^(2)` so `m_(p)v_(p) =sqrt(2m_(p)E)`…..(i) For ELECTRON K.E. `E_(e) =1/2m_(e)v_(e)^(2)`, So `m_(e)v_(e)=sqrt(2m_(e)E)`……(ii) So the ratio of de-Broglie wavelength of PROTONS and ELECTRONS is `lambda_(p)/lambda_(e) =h/(m_(p)v_(p)) xx (m_(e) v_(e))/h = (m_(e) v_(e))/(m_(p)v_(p))` Using Eq. (i) & (ii), we get `lambda_(p)/lambda_(e) =sqrt((2m_(e)E)/(2m_(p)E)) = sqrt(m_(e)/m_(p)) lt 1` `lambda_(p) lt lambda_(e)` or `lambda_(e) gt lambda_(p)` .e. de-Broglie wavelength associated with electron is more than that of proton. |
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