Question

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

Answer 1

Answer: (D)

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
$K.E$ photoelectron $=hv-\varphi$
Stopping potential is to stop the fastest photoelectron
$V_0=\frac{hv}{e}-\frac{\varphi }{e}$
so, $K.E_{max}$ and $V_0$ both increases.
But $K.E$ ranges from zero to K. $E_{max}$ because of loss of energy due to subsequent collisions before getting ejected and not due to range of frequencies in the incident light.

So, the answer is option (D).