Question

Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by X-rays, both $V_0$ and $K_{max}$ increase.
Statement-2 : photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

• A. Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1.
• B. Statement-1 is true, Statement-2 is true: Statement-2 is not the correct explanation of Statement-1
• C. Statement-1 is false, Statement-2 is true.
• D. Statement-1 is true, Statement-2 is false.

Answer 1

The correct option is D Statement-1 is true, Statement-2 is false.

Since the frequency of ultraviolet light is less than the frequency of X-rays, the

energy of each incident photon will be more for X-rays

KEphotoelectron=hv–ϕ

Stopping potential is to stop the fastest photoelectron

V∘=$\frac{hv}{e}–\frac{\varphi }{e}$

and V∘ both increases. So, KEmax

But KE ranges from zero to KEmax because of loss of energy

due to subsequent collisions before getting ejected and not due to

range of frequencies in the incident light.

hence option D is the correct answer