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An electromagnetic wave of wavelength lamda is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength lamda_(1), prove that lamda=((2mc)/(h))lamda_(1)^(2). |
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Answer» Solution :As per Einstein.s PHOTOELECTRIC equation, we know that `(hc)/(lamda)-phi_(0)=K` If WORK function `phi_(0)` is negligible, then `(hc)/(lamda)=K` where K is the maximum vlaue of KINETIC energy of ejected photoelectrons If de-Broglie WAVELENGTH of ejected photoelectrons be `lamda_(1)`, then `lamda_(1)=(h)/(p)=(h)/(SQRT(2mK))implies K=(h^(2))/(2mlamda_(1)^(2))` Comparing two expressions for K, we have `(hc)/(lamda)=(h^(2))/(2mlamda_(1)^(2))implies lamda=((2mc)/(h))lamda_(1)^(2)`. |
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