Light rays consisting of photons with energy $1.8 \ eV$ are incident on a metal surface with a work function of $1.2 \ eV$. What is the stopping potential required to stop the emitted electrons in $eV$?

The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

The given figure shows a few data points in a photoelectric effect experiment for a certain metal. The minimum energy for the ejection of an electron from its surface is $....... eV$. (Planck's constant $h = 6.62 \times 10^{-34} \, J \cdot s$)

The work functions for the following metals are given:
$Na: 2.75\;eV; K: 2.30\;eV; Mo: 4.17\;eV; Ni: 5.15\;eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\,\mathring{A}$ from a $He-Cd$ laser placed $1\;m$ away from the photocell? What happens if the laser is brought nearer and placed $50\;cm$ away?

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is: