When the energy of incident radiation is increased by $20 \%$,the kinetic energy of the photoelectrons emitted from a metal surface increases from $0.5 \ eV$ to $0.8 \ eV$. The work function of the metal is (in $eV$)

The work function of a metal is $2.3 \ eV$ and the wavelength of the incident photon is $4.84 \times 10^{-7} \ m$. What is the maximum kinetic energy of the photoelectrons in $eV$?

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

Electrons ejected from the surface of a metal,when light of a certain frequency is incident on it,are stopped fully by a retarding potential of $3 \ V$. The photoelectric effect on this metallic surface begins at a frequency of $6 \times 10^{14} \ s^{-1}$. The frequency of the incident light in $s^{-1}$ is: [Planck's constant $= 6.4 \times 10^{-34} \ J \cdot s$,charge on the electron $= 1.6 \times 10^{-19} \ C$]

$A$ photon of $5.5 \ eV$ energy falls on the surface of a metal,emitting photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping voltage required for these electrons is .......... $V$.

$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)