The frequency of incident light falling on a photo sensitive plate is doubled, then maximum kinetic energy of the emitted photoelectrons will become –
Answer & explanation
Correct answer: option 2
$KE_{max}=h(v-v_0)$
$\frac{KE_{max}}{KE_{max}}=\frac{h(2v-v_0)}{h(v-v_0)}>2$