Photons of energy $10 eV$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} Hz$. The kinetic energy in $eV$ of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34} Js$]. (in $eV$)

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

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

When a point source of light is at a distance of $1 \ m$ from a photocell,the cut-off voltage is found to be $V$. If the same source is placed at $2 \ m$ distance from the photocell,the cut-off voltage will be

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 $K_0$)

When a photon of energy $6 \, eV$ is incident on a metal surface,the maximum kinetic energy of the emitted photoelectrons is $4 \, eV$. The value of the stopping potential is .......... $V$.