Photoelectric Effect

The photoelectric effect is a phenomenon where metals emit electrically charged particles upon absorbing electromagnetic radiation. Specifically, it involves the ejection of electrons from a metal surface. This effect is the foundation of modern physics as it addresses a fundamental question about the nature of light. Albert Einstein famously explained this phenomenon in 1905. However, Heinrich Rudolf Hertz laid the groundwork for the photoelectric effect theory in 1887 through his early research and discoveries. When light of a specific frequency strikes a metal surface, it causes the emission of electrons, a phenomenon known as the photoelectric effect. These emitted electrons are referred to as photoelectrons. The emission generates an electric current known as a photocurrent. The photoelectric effect is pivotal in demonstrating the quantum nature of radiation. Classical electromagnetic theory, however, needed a satisfactory explanation for this phenomenon.

1.0Definition of Photoelectric Effect

It is a phenomenon of ejecting electrons by falling light of a suitable frequency or wavelength on a metal. Ejected electrons are called photoelectrons, and the current flowing due to the photoelectrons is called photoelectric current. This effect was discovered by Hertz. Lenard gave the laws of the photoelectric effect. Einstein explained it using the Quantum theory of light. 

2.0Photoelectric Effect Uses

• Solar cell Panel for Electricity Generation.
• To control the Temperature of the Furnace in the industry.
• Photoelectric Sensors.
• Photoelectric Smoke Detectors.

3.0Photoelectric applications:

• Optical Fiber used in Telecommunication
• Burglar and Fire Alarms
• Television camera for Telecasting scenes
• The Study of the Spectrum of Stars and Temperature.

4.0Photoelectric Effect Work Function

• It is the minimum energy an electron requires to escape from a metal surface.
• Work function depends on the nature of the metal and its surface.
• Minimum Work Function: Caesium ( Cs) = 2.14 eV.
•  Maximum Work Function: Platinum ( Pt )= 5.65 eV.  (∵ 1eV = 1.6 ×10^–19 J)

5.0Experimental Set Up of Photoelectric Effect

• This experimental setup consists of an evacuated glass tube in a vacuum with two electrodes, A and C. The electrode C (called cathode) is a photo-sensitive emitter that emits photoelectrons when exposed to ultraviolet radiation. The electrode A (called the anode) is a charge-collecting plate. The tube has a side window of quartz covered with a filter through which the incident light of the desired wavelength enters the tube and falls on the photosensitive plate C.
• Electrodes A and C are connected to a battery through a suitable reversing switch, S .This switch helps bring Electrode A to a positive or negative potential relative to Electrode C.
• Electrons are emitted when ultraviolet radiations are made to fall on photosensitive plate C. These electrons get attracted towards electrode A when it has a positive potential relative to electrode C. The flow of electrons from electrode C to electrode A makes electric current flow in the circuit. This electric current is known as photoelectric current because of the photoelectrons emitted by photo-sensitive electrode C. The photoelectric current is measured with a microammeter (A) connected in series. The potential difference between plates A and C is measured with the help of a voltmeter (V).

6.0Explanation of Photoelectric Effect

When a constant potential difference is applied across electrode A and electrode C, and when ultraviolet light is incident on cathode C, photoelectrons are emitted, which are controlled by plate or anode A. The photoelectric current (I_P) due to these photoelectrons are measured with a microammeter(A). As the intensity of incident light increases, keeping frequency constant, more and more photoelectrons are emitted by electrode C and photoelectric current increases linearly. The variation of photoelectric current with the intensity of light is linear.

• Photoelectric current (I_P) is directly proportional to the number of photoelectrons emitted per second.
•  The no. of electrons emitted per second is directly proportional to the intensity of incident light, provided the frequency of incident light is greater than the threshold frequency.

Threshold Frequency

The minimum frequency of incident light required to emit electrons from a metal surface is known as the threshold frequency. Photoelectrons are emitted from an emitter (metal) surface only if the incident light frequency is greater than or equal to the threshold frequency.

Stopping Potential

Minimum magnitude of negative potential of anode with respect to cathode for which current is zero is called stopping potential. This is also known as cutoff voltage. This voltage is independent of intensity.

Retarding potential

Negative potential of anode with respect to cathode which is less than stopping potential is called retarding potential

7.0Laws of Photoelectric Emission

1)  For a given substance, there is a minimum frequency of incident light called threshold frequency( 𝜈0) below which no photoelectric emission takes place whatsoever the intensity of incident light may be.

(2)  The number of photoelectrons emitted per second( photoelectric current) by a substance is directly proportional to the intensity of incident light.

(3) The maximum kinetic energy of the photoelectrons increases with the increase in the frequency of the incident light, provided the frequency of incident light is greater than the threshold frequency.

(4) The process of photoelectric emission is instantaneous.

8.0Einstein-Photoelectric Equation

As per Einstein, an incident photon with energy (h𝜈) ejects an electron from a metal surface with a work function( 0) and imparts its kinetic energy, $\text{ K.E }=\frac{1}{2}m{v}^2$. According to Einstein, when a photon of energy (h𝜈) falls on a metal surface, the energy of the photon is used in two ways:

(1). An electron uses some energy electrons to cross the surface barrier so that it may come out of the metal surface. This part of the energy is equal to the work  function of the metal.

(2). The remaining energy is used to give velocity to the emitted photoelectrons. This part of the energy equals the kinetic energy of emitted photoelectrons.

According to the Law of Conservation of Energy,

$\mathrm{h}v={\varphi }_0+\frac{1}{2}{\mathrm{mv}}^2$………......….(1)

if 𝜈 is just equal to the threshold frequency 0 then the free electron emitted from the metal surface will not have any kinetic energy ($\frac{1}{2}m{v}^2=0$)

Hence equation (1) can be written as

$\mathrm{h}{v}_0={\varphi }_0+0$

${\varphi }_0=\mathrm{h}{v}_0$ …………….(2)

Substituting equation equation ( 2) in (1), we get

$\mathrm{h}v=\mathrm{h}{v}_0+\frac{1}{2}{\mathrm{mv}}^2$

$\mathrm{h}\left(v-v_0\right)=\frac{1}{2}{\mathrm{mv}}^2$

$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{h}\left(v-v_0\right)$

$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{h}\left(v-v_0\right)$..........Einstein Photoelectric equation

9.0Graphs of Photoelectric Effect

Effect of Intensity of incident light on the photoelectric current

Effect of potential on Photoelectric Current

Effect of Frequency of Incident Light

Graph for Frequency and Stopping Potential

The Graph between Kinetic Energy and Frequency

10.0Solved Problems

Q-1. Two beams, one of red light and one of blue light, having the same intensity, are incident on a metal plate to do the emission of electrons. Which light imparts more kinetic energy?

Sol. $\text{ K.E }\propto v\text{ (Frequency) }\propto \frac{1}{\lambda }$

${\lambda }_{\text{blue light }}$is less than that of Red light.So Blue light emits electrons of greater K.E.

Q-2. The work function of Na is 2.3 eV. Does Sodium show photoelectric emission for light of wavelength 6800 Å?

Sol. $\mathrm{E}=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{6800\times 10^{-10}}=2.9\times 10^{-19}$

$\mathrm{J}=\frac{2.9\times 10^{-19}}{1.6\times 10^{-19}}\mathrm{eV}=1.8\mathrm{eV}$

EくWork Function so Sodium will not show photoemission.

Q-3. Removing a free electron from Copper than from Sodium is more difficult. Why?

Sol. Work function ${\varphi }_0=h{\nu }_0=\frac{hc}{{\lambda }_0}$

 ${\lambda }_0$= Threshold Wavelength

${\lambda }_{ONa}>{\lambda }_{OCu}$ ( Work Function of copper is greater than that of Sodium)

Hence, the Threshold wavelength for Sodium is greater than that of Copper.