Question

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is,
A. 2
B. 4
C. 6
D. 10

Answer 1

Hint: Photoelectrons are emitted from a metal, when illuminated with light, above the threshold frequency of the metal, with a range of KE’s. The stopping potential is the voltage between the metal surface and a cathode, placed close to the surface, in a vacuum, which just stops them from producing a current. We can then assume that the max KE of the photoelectrons, ${{K}_{\max }}=e.{{V}_{{}^\circ }}$
where ${{V}_{{}^\circ }}$ is the stopping potential.

Complete answer:
Given, the maximum kinetic energy: ${{K}_{\max }}=4e.V$
 If ${{V}_{{}^\circ }}$ be the stopping potential, then ${{K}_{\max }}=e.{{V}_{{}^\circ }}$

$\begin{align}
  & =e.{{V}_{{}^\circ }}=4e.V \\
 & ={{V}_{{}^\circ }}=4V \\
\end{align}$

The stopping potential in volts required is 4 volts.

Hence the correct option is C.

Additional information:
We know that the stopping voltage is an opposing force that prevents the charges from moving in the presence of the photons. So basically, stopping voltage can also be described as the equivalent emf that prevents charges from moving which are energized by photons. Because it is exactly equal to the photoelectric force applied, we can use the calculated stopping voltage to calculate the energy level of the electrons.
Its significance is that it can be also used to determine the energy level of the energized electrons.
We can say that it is a value that clearly depends on the maximum kinetic energy that a photo electron may have. This, in turn, depends on the frequency of the incident radiation and work function of the irradiated surface. We know that intensity gives a measure of the number of photoelectrons emitted, not a measure of how fast they move after emission.

Note:
When light of suitable frequency V (greater than threshold frequency ${{V}_{{}^\circ }}$) or suitable wavelength (lesser than threshold wavelength) is emitted on a suitable metal target (of work function lesser than threshold energy of light used)
Then photo-electrons are emitted from metal having kinetic energy ranging from zero to ${{K}_{\max }}$
${{K}_{\max }}$ is equal to $h\nu -h{{\nu }_{{}^\circ }}$
To stop the photo-electrons at the surface of metal, we need to use a positive plate of minimum potential difference, say ${{V}_{{}^\circ }}$
Then energy applied by the electric field on electron is $e.{{V}_{{}^\circ }}$ for minimum potential difference, Vs is called stopping potential and then
${{K}_{\max }}=e.{{V}_{{}^\circ }}$