If p is the momentum of the fastest electron | vedclass

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(C) The kinetic energy $(K.E.)$ of the photoelectron is given by $K.E. = \frac{p^2}{2m}$.According to the photoelectric equation,$\frac{hc}{\lambda} =

Vedclass · Accepted Answer

$\frac{4}{9} \lambda$

Vedclass · Answer

(C) The kinetic energy $(K.E.)$ of the photoelectron is given by $K.E. = \frac{p^2}{2m}$.According to the photoelectric equation,$\frac{hc}{\lambda} = W_o + K.E.$Given that $K.E. \gg W_o$,we can approximate $\frac{hc}{\lambda} \approx K.E. = \frac{p^2}{2m}$.For the first case,$\frac{hc}{\lambda} = \frac{p^2}{2m}$.For the second case,let the new wavelength be $\lambda'$ and momentum be $p' = 1.5 \ p = \frac{3}{2} \ p$.Then,$\frac{hc}{\lambda'} = \frac{(1.5 \ p)^2}{2m} = \frac{2.25 \ p^2}{2m}$.Dividing the two equations: $\frac{\lambda}{\lambda'} = \frac{2.25 \ p^2 / 2m}{p^2 / 2m} = 2.25 = \frac{9}{4}$.Therefore,$\lambda' = \frac{4}{9} \lambda$.

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