The correct graph between the maximum kinetic energy $(K_{\max})$ of a photoelectron and the inverse of the wavelength $(1/\lambda)$ of the incident radiation is given by the curve:

For intensity $I$ of a light of wavelength $5000 \, Å$, the photoelectron saturation current is $0.40 \, μA$ and the stopping potential is $1.36 \, V$. The work function of the metal is ........... $eV$.

The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)

The work function of a metal is $3 \ eV$. The color of the visible light that is required to cause emission of photoelectrons is

When photons of energy $hf$ fall on an aluminium plate (of work function $E_0$),photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled,the maximum kinetic energy of the ejected photoelectrons will be:

Consider a metal exposed to light of wavelength $600 \, nm$. The maximum kinetic energy of the emitted electron doubles when light of wavelength $400 \, nm$ is used. Find the work function of the metal in $eV$. (in $, eV$)