When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

The work functions of metals $A$ and $B$ are in the ratio $1 : 2$. If light of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energy of photoelectrons emitted will be ($f$ and $2f$ are both frequencies greater than the threshold frequency of metals $A$ and $B$).

Light of frequency $v$ is incident on a certain photoelectric material with a threshold frequency $v_0$. What is the work function for the material?

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:

When the energy of the 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 ............. $eV$.

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by: