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Kinetic energy remaining the same, which of the two have greater de-Broglie wavelength proton or electron ? Explain. |
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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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