The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when light of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then

When a point source of light is at a distance of $50 \, cm$ from a photoelectric cell,the cut-off voltage is found to be $V_0$. If the same source is placed at a distance of $1 \, m$ from the cell,then the cut-off voltage will be

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

An electron moving with velocity $1.6 \times 10^7 \ m/s$ has a wavelength of $0.4 \ \mathring{A}$. The required accelerating voltage for the electron motion is (charge on electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$).

The threshold frequency of a photosensitive material is $v$. When photons of frequency $2v$ are incident on the material,photoelectrons are emitted with a maximum linear momentum $P$. To get photoelectrons with maximum linear momentum $2P$,the frequency of the incident photons is: (in $v$)

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.