When light falls on a metal surface,the maximum kinetic energy of the emitted photo-electrons depends upon

$A$ photon of energy $8\,eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15}\,Hz$. The maximum kinetic energy of photoelectrons emitted is .......... $eV$.
(Take $h = 6.6 \times 10^{-34}\,J\cdot s$; $1\,eV = 1.6 \times 10^{-19}\,J$)

The threshold wavelength for photoelectric emission in tungsten is $400 \ nm$. The wavelength of light that must be used in order to eject electrons with a maximum kinetic energy of $0.9 \ eV$ is .............. $nm$.

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda/3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case, the work function of the surface of the material is ($h$ = Planck's constant, $c$ = speed of light).

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be