When photons of energy h fall on an aluminium plate (of work function E0), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be
- K + h
- K + E0
- 2K
- K.
Answer
(1) K + h
Solution
K = h - E0
For frequency 2 : K' = 2h - E0
K' = (h - E0) + h = K + h
Answer: (1)