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When photons of energy (hc)/(lamda) fall...

When photons of energy `(hc)/(lamda)` fall on a metal surface, photoelectrons are ejected from it. If the work function of the surfacei s `hupsilon_0`, then

A

maximum kinetic energy of the electron is `((hc)/(lambda)-hv_(@))`

B

maximum kinetic energy of the photoelectron is equal to `(hc//lambda)`

C

minimum `K.E.` of photoelectron is zero.

D

minimum kinetic energy of the photoelectron is equal to `hc//lambda`

Text Solution

Verified by Experts

Use Einstein's photoelectric equation
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