When radiation of frequency $8 \times 10^{15} \ Hz$ is incident on a metal surface with a work function of $6.125 \ eV$,what is the kinetic energy of the emitted photoelectrons in $eV$?

The variation of photo-current with collector potential for different frequencies of incident radiation $v_{1}, v_{2}$ and $v_{3}$ is as shown in the graph. Then:

The work function of a metal is $6.825 \ eV$. Its threshold wavelength is approximately: (Given $c = 3 \times 10^8 \ m/s$)

Energy required to remove an electron from an aluminum surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ Å}$ falls on the surface,the velocity of the fastest electron ejected from the surface will be:

If a photocell is illuminated with a radiation of $1240 \mathring{A}$,the stopping potential is found to be $8 \text{ V}$. Then the work function of the emitter and the threshold wavelength are:

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: