The light rays having photons of energy $4.2\,eV$ are falling on a metal surface having a work function of $2.2\,eV$. The stopping potential of the surface is $.........\,V$.

The variation of stopping potential $V_{0}$ with frequency $\nu$ of incident radiation for a given photosensitive material is a straight line [frequency $\nu$ of incident radiation is greater than threshold frequency $\nu_{0}$]. The slope of this line is . . . . . . .

Five elements $A, B, C, D$ and $E$ have work functions $1.2 \, eV, 2.4 \, eV, 3.6 \, eV, 4.8 \, eV$ and $6 \, eV$ respectively. If light of wavelength $4000 \, Å$ is allowed to fall on these elements, then photoelectrons are emitted by:

The photoelectric cut-off voltage in a certain experiment is $1.5 \ V$. What is the maximum kinetic energy of the emitted photoelectrons?

$A$ photon of energy $4 eV$ imparts all its energy to an electron that leaves a metal surface with $1.1 eV$ of kinetic energy. The work function of the metal is (in $eV$)

In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)