Question

Cut off potentials for a metal in photoelectric effect for light of wavelength $\lambda_{1}$ , $\lambda_{2}$ and $\lambda_{3}$ is found to be $V_{1}$, $V_{2}$ and $V_{3}$ volts if $V_{1}$, $V_{2}$ and $V_{3}$ are in Arithmetic Progression and $\lambda_{1}$ , $\lambda_{2}$ and $\lambda_{3}$ will be:
A) Arithmetic Progression
B) Geometric Progression
C) Harmonic Progression
D) None of these

Answer 1

Hint: The cut-off potential is represented as the necessary potential for stopping the departure of an electron from a metal surface while the energy of the incident light is greater than the work potential of the metal at which the incident light is concentrated. A harmonic progression is a progression made by taking the reciprocals of an AP. Equivalently, a series is a harmonic progression when every term is the harmonic average of the adjacent terms.

Complete step-by-step solution:
Kinetic Energy of the electrons, $K=\dfrac{hc}{\lambda}$
This Kinetic energy is equal to the stopping potential.
$eV =\dfrac{hc}{\lambda}$
This gives,
$V =\dfrac{hc}{e\lambda}$ .
Hence,
$V_{1} =\dfrac{hc}{e\lambda_{1}}$
$V_{2} =\dfrac{hc}{e\lambda_{2}}$
$V_{3} =\dfrac{hc}{e\lambda_{3}}$
$V_{1}$, $V_{2}$ and $V_{3}$ are in Arithmetic Progression.
Therefore, $V_{1} – V_{2} = d = V_{2} – V_{3}$
d is the successive difference.
$\dfrac{hc}{e\lambda_{1}} - \dfrac{hc}{e\lambda_{2}}= \dfrac{hc}{e\lambda_{2}}-\dfrac{hc}{e\lambda_{3}}$
This gives,
$\dfrac{1}{\lambda_{1}} - \dfrac{1}{\lambda_{2}}= \dfrac{1}{\lambda_{2}}-\dfrac{1}{\lambda_{3}}$
Hence, $\lambda_{1}$ , $\lambda_{2}$ and $\lambda_{3}$ will be Harmonic Progression.
Option (C) is correct.

 Note: Stopping potential is independent of the incident radiation intensity. On rising intensity, the value of saturated current increments, whereas the stopping potential continues unchanged. For a provided intensity of radiation, the stopping potential depends on the frequency—the larger the frequency of incident light greater the value of stopping potential.