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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function

Vedclass · Accepted Answer

$\frac{\lambda}{2} < \lambda' < \lambda$

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(C) According to Einstein's photoelectric equation,the kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - W_0$,where $W_0$ is the work function.For kinetic energy $2E$,the new wavelength $\lambda'$ satisfies: $2E = \frac{hc}{\lambda'} - W_0$.From the first equation,$\frac{hc}{\lambda} = E + W_0$,and from the second,$\frac{hc}{\lambda'} = 2E + W_0$.Dividing the two equations: $\frac{\lambda'}{\lambda} = \frac{E + W_0}{2E + W_0}$.This can be written as: $\lambda' = \lambda \left( \frac{1 + W_0/E}{2 + W_0/E} \right)$.Since $W_0 > 0$,the term $\frac{1 + W_0/E}{2 + W_0/E}$ is always greater than $\frac{1}{2}$ and less than $1$.Therefore,$\frac{\lambda}{2} < \lambda' < \lambda$.

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where

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