The kinetic energy of photo electron depends on
A) Frequency of incident radiation
B) Velocity of incident radiation
C) Intensity of incident radiation
D)Wavelength of incident radiation

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\mathrm{h}(\mathrm{\nu }-{\mathrm{\nu }}_0) \\ \mathrm{where}, \mathrm{\nu }=\mathrm{frequency} \mathrm{of} \mathrm{incident} \mathrm{light} \\ {\mathrm{\nu }}_0=\mathrm{Threshold} \mathrm{frequency} \end{gathered}$$

From this,

As frequency of incident radiation increases, K.E increases

As wavelength of incident radiation increases, K.E decreases  $\left(\because \nu \propto \frac{1}{\lambda }\right)$

Intensity of incident radiation increases, no. of photo electrons emitted increases but not influences K.E.