When the frequency of the incident radiation on a metallic plate is doubled, the kinetic energy of the photoelectrons will be:

doubled

No worries! We‘ve got your back. Try BYJU‘S free classes today!

halved

No worries! We‘ve got your back. Try BYJU‘S free classes today!

more than doubled

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

increases but less than doubled

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C more than doubled
$K E_{1} = h ν - W_{0}$
$K E_{2} = 2 h ν - W_{0}$
If $K E_{2}$ were equal to $(2 h ν - 2 W_{0}),$ then $K E_{2}$ would have been doubled.
But, here $K E_{2} = 2 h ν - W_{0}$
i.e., $(2 h ν - W_{0}) > (2 h ν - 2 W_{0})$
So, the kinetic energy of the photoelectrons will be more than doubled.

Suggest Corrections

Similar questions

E

ν

E

ν

The frequency of incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

Join BYJU'S Learning Program