#### Question

If the frequency of incident light falling on a photosensitive material is doubled, then the kinetic energy of the emitted photoelectron will‘be........................................

- same as its initial value
- two times its initial value
- more than two times its initial value
- less than two times its initial value

#### Solution

(c) more than two times its initial value

Since the frequency is double, the value of E is also double.

K.E. = E - Φ

∴ K.E._{1} = E - Φ ............(i)

K.E._{2} = 2E - Φ ..............(ii)

From (i) and (ii), we get

K.E._{2} > K.E._{1}

Is there an error in this question or solution?

#### APPEARS IN

Solution If the Frequency of Incident Light Falling on a Photosensitive Material is Doubled, Then the Kinetic Energy of the Emitted Photoelectron Will‘Be Concept: K.E.(Kinetic Energy) and P.E.(Potential Energy) in S.H.M..