Question

Statement -1 :When ultraviolet light is incident on a photo cell, its potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultravoilet light is replaced by X-rays, both $V_0$ and $K_{max}$ increase.

Statement-2: Photoelctrons 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 correct explanation of statement-1.
• B. Statement -1 is true,statement-2 is true ; statement -2 is not 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

Given: When ultraviolet light is incident on a photo cell, potential= V0 and maximum K.E.=Kmax. When X- rays is incident on a photo cell, both V0 and Kmax increase.

Solution: As we know that the frequency of X-rays is more than that of uv-rays

Then K.E. of photoelectron in X-rays will be greater than in UV- rays(\because K.E.=h\nu -\phi)

And stopping potential in X-rays will be greater than in UV- rays(\because V_{0}=\frac{h\nu }{e}-\frac{\phi }{e})

So, statement 1 is correct.

Photoelctrons are emitted with speeds ranging from zero to a maximum value because of the collisions in ejecting electrons there is loss of enegy but not due to range of frequencies present in the incident light.

Statement 2 is wrong.

Hence the correct answer is D