A metal plate is exposed to light with wavelength 5000 nm. It is observed that electrons are ejected from the surface of the plate. When a retarding uniform electric field 0.01 N/C. is imposed, no electron can move away from the plate farther than a certain distance d. If the threshold wavelength is twice the incident light wavelength for the material of the plate then the distance d is _______ m. (Use $h c = 12400 e V \overset{0}{A}$ )

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answer is 12.4.

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Detailed Solution

The maximum kinetic energy of the electrons immediately upon ejection is the difference between the energy of the incident photon and the threshold energy.
$K = \frac{h c}{λ} - \frac{h c}{λ_{0}}$
This kinetic energy of ejected electron is converted to electrostatic potential energy, $Δ U = e E d$ , as electrons come to rest while moving in the direction of electric field. Therefore, $K = E e d$ . and $λ_{0} = {(\frac{1}{λ} - \frac{e E d}{h c})}^{- 1} \Rightarrow λ_{0} = 2 λ \Rightarrow d = \frac{h c}{2 λ e E} = 12.43 m$

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