The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is:

The photoelectric threshold wavelength for a certain metal surface is $3300 \,\mathring{A}$. What is the maximum kinetic energy of the photoelectrons emitted if radiations of wavelength $1100 \,\mathring{A}$ are used?

The beam of light has three wavelengths $4144 \,\mathring A$,$4972 \,\mathring A$,and $6216 \,\mathring A$ with a total intensity of $3.6 \times 10^{-5} \, W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area of $1 \, cm^2$ of a clean metallic surface with a work function of $2.3 \, eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \, s$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively 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 $h = 6.62 \times 10^{-34} \, J \cdot s$)