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Q. 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.

AIEEEAIEEE 2010Dual Nature of Radiation and Matter

Solution:

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
$K.E _{photoelectron} = hn -\varphi$
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.