Question

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work functions $5.01eV$, when ultraviolet light $200nm$ falls on it must be.
A. $2.4V$
B. $1.2V$
C. $- 1.2V$
D. $- 2.4V$

Answer 1

The electron is emitted from the surface of the metal when the light with enough frequency incident on the metal surface. The energy is distributed uniformly across the wavefront and it only depends on the beam’s intensity.
When the light intensity gets increased then the kinetic energy of the electrons gets increased.

Complete step by step solution:
When the frequency is larger than the threshold frequency then the electrons get emitted. Which is based on the wave theory. Energy fully depends on the intensity of light.
The following equation explains the photoelectric effect,
$E = hf$
$E$ is the energy required.
$h$ is the Plank constant
$f$ is the frequency.
Plank constant explains the behavior of the atomic particles. This plank constant is also used to understand the photoelectric effect and quantum mechanics.
The energy of incident light is $\dfrac{{12375}}{{2000}}$
${V_0} = 1.2V$
$E = 6.2eV$
By using the relation,
$E = {W_0} + e{V_0}$
From this, the ${V_0}$
${V_0} = \dfrac{{E - {W_0}}}{e}$
Now apply the values in the above equation,
${V_0} = \dfrac{{(6.2 - 5.01)e}}{e}$
Now solve the above equation the ${V_0}$ is,
${V_0} = 1.2V$
So, the correct answer is option (B) ${V_0} = 1.2V$.

Note:
When the photon lies on the metal surface the energy of the photon energy is transferred to the electron.
The electrons emitted from under the metal surface lose the kinetic energy because of the collision.
The kinetic energy is carried by the surface electrons which are imparted by the photon with the maximum energy.