Photoelectron are emitted with maximum kinetic energy E from a metal surface when light of frequency v falls on it when light of frequency v' falls on the same metal, the max. KE. Of emitted Photoelectrons is found to be 2E then v' is -
Answer & explanation
Correct answer: option 3
$KE = hv + \phi$ …..(i)
$2KE = hv' + \phi$ ……(ii)
or $2 (hv + \phi) = hv' + \phi$
or $v' = 2v +\frac{\phi}{h}⇒v' > 2v$