Radiations of wavelength $400 \,nm$ are incident on a photosensitive material with a work function of $2.2 \,eV$. The stopping potential is nearly: (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.

All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

$A$ photosensitive metallic surface is illuminated alternately with lights of wavelength $3100 \mathring A$ and $6200 \mathring A$. It is observed that the maximum speeds of the photoelectrons in the two cases are in the ratio $2:1$. The work function of the metal is $(hc = 12400 \, eV \mathring A)$.