Question

It is observed that photoelectric emission stops at a certain time t after the light source is switched on. It is due to the retarding potential developed in the metallic sphere due to leftover positive charges. The stopping potential (V) can be represented as,
(A). $2\left( K{{E}_{\max }}/e \right)$
(B). $\left( K{{E}_{\max }}/e \right)$
(C). $\left( K{{E}_{\max }}/3e \right)$
(D). $\left( K{{E}_{\max }}/2e \right)$

Answer 1

Hint: Study about the wave nature and the particle nature of light. Imagine how they are different. Try to solve this photoelectric effect using both wave and particle nature. Only the particle nature of light is suitable for this phenomenon. Obtain the equation using the concepts of photoelectric effect.

Complete step by step answer:
Photoelectric effect is a phenomenon where electrons from the surface of a metal are emitted by absorbing light.
When a light of sufficient energy is incident on a metal surface, the electrons on the surface of the metals absorb this light and emitted out of the surface of the metal.
According to the wave theory of light, light is a form of wave which propagates through space. But it cannot explain the photoelectric effect.
Einstein introduced the particle theory of light. Where he explained that light is a particle called a photon. The energy carried by each particle of light called photon depends on the frequency as,
$E=h\nu $
Where h is the Planck’s constant with value, $h=6.626\times {{10}^{-34}}Js$
When a photon with sufficient energy is incident on the metal surface the electron absorbs the energy. With energy required to eject the electron. The electron is emitted and the rest of the energy is converted to the kinetic energy of the electron.
So, we can write,
$\text{energy of photon = energy required to eject the electron + kinetic energy of the ejected electron}$
$E=W+K.E.$
Where, W is the work function or the threshold energy below which photoelectric effect is not possible
$E=h{{\nu }_{0}}+\dfrac{1}{2}{{m}_{e}}{{v}^{2}}$
The kinetic energy of the photoelectron is given by,
$K.E.=E-h{{\nu }_{0}}$
Where, ${{\nu }_{0}}$ is the threshold frequency.
Stopping potential of the photoelectron is equal to its kinetic energy.
So, we can write that $K.E{{.}_{\max }}=eV$
Where V is the stopping potential.
So, we can write that,
$V=\dfrac{K.E{{.}_{\max }}}{e}$
So, the correct option is (A).

Note: The opposite of photoelectric effect is also possible. When we have a beam of fast-moving electrons near any positively charged metal surface, it can absorb the electron emitting light of certain wavelength.