Introduction and Photoelectric Effect

Emission of electrons

At room temperature the free electrons move randomly within the conductor, but they don't leave the surface of the conductor due to attraction of positive charges. Some external energy is required to emit electrons from a metal surface. Minimum energy is required to emit the electrons which are just on the surface of the conductor. This minimum energy is called the work function( W) . The work function is the property of the metallic surface.

The energy required to liberate an electron from metal surface may arise from various source such as heat, light, electric field etc. Depending on the nature of source of energy, the following methods are possible.

(i) Thermionic emission: The energy to the free electrons can be given by heating the metal, the electrons so emitted are known as thermions.

(ii) Field emission: When a conductor is put under strong electric field the free electrons on it experience an electric force in the opposite direction of field. Beyond a certain limit electrons start coming out of the metal surface. Emission from a metal surface by this method is called the field emission.

(iii) Secondary emission: Emission of electrons from a metal surface by the bombardment of high speed electrons or other particle is known as secondary emission.

(iv) Photoelectric emission: Emission of free electrons from a metal surface by falling light (or any other electromagnetic wave which has an energy greater than the work function of the metal) is called photoelectric emission. The electrons so emitted are called photoelectrons. This is explained in detail as under.

PHOTOELECTRIC EFFECT

It was observed by Hertz and Lenard around 1880 that when a clean metallic surface is irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomenon of ejection of the electrons from metal surface was called as Photoelectric Effect. The electrons thus ejected were called as photoelectrons. For photoemission to take place, energy of incident light photons should be greater than or equal to the work function of the metal.

Or E $⩾$ W

hf $⩾$ W [Where h is plank's constant]

f $⩾$ $\frac{W}{h}$

Here $\frac{W}{h}$ is the minimum frequency required for the emission of electrons. This is known as threshold frequency f_o.

Study of photoelectric effect

The given set up (as shown in fig.) is used to study the photoelectric effect experimentally.

In an evacuated glass tube, two zinc plates C and D are enclosed. Plates C acts as anode and D acts as photosensitive plate. Two plates are connected to a battery B and ammeter A. If the radiation is incident on the plate D through a quartz window W, electrons are ejected out of plate and current flows in the circuit. The plate C can be maintained at desired potential (+ve or -ve) with respect to plate D. With the help of this apparatus, we will now study the dependence of the photoelectric effect on the following factors.

1. Intensity of incident radiation

2. Potential difference between C and D

3. Frequency of incident radiation.

Effect of Intensity of incident radiation

The electrode C i.e. collecting electrode is made positive with respect to D. Keeping the frequency of light and the potentials fixed, the intensity (amount of energy falling per unit area per second) of incident light is varied and the photoelectric current (i) is measured in ammeter. The photoelectric current is directly proportional to the intensity of light. The photoelectric current gives an account of number of photoelectrons ejected per sec.

Effect of p.d. between C & D

Keeping the intensity and frequency of light constant, the positive potential of C is increased gradually. The photoelectric current increases with increase in voltage (accelerating voltage) till, for a certain positive potential of plate C, the current becomes maximum beyond which it does not increase for any increase in the accelerating voltage. This maximum value of the current is called as saturation current.

$\bullet$ Make the potential of C as zero and make it increasingly negative. The photoelectric current decrease as the potential is made increasingly negative (retarding potential), till for a sharply defined negative potential V_c of C, the current becomes zero. The retarding potential for which the photoelectric current becomes zero is called as cut-off or stopping potential (V_c).

$e{V}_C=\frac{1}{2}m{v}_{\max }^2$

When light of same frequency is used at higher intensity, the value of saturation current is found to be greater, but the stopping potential remains the same. Hence the stopping potential is independent of intensity of incident light of same frequency.

Effect of frequency on Photoelectric Effect

$\bullet$ The stopping potential V_c is found to be changing linearly with frequency of incident light being more negative for high frequency. An increase in frequency of the incident light increases the kinetic energy of the emitted electrons, so greater retarding potential is required to stop them completely. For a given frequency v, V_c measures the maximum kinetic energy E_max of photoelectrons that can reach plate C.

$\Rightarrow e{V}_c=\frac{1}{2}m{V}_{\max }^2$

Where m is the mass of electron, e is charge of electron and V_max is maximum velocity of electron. This means that work done by stopping potential must just be equal to maximum kinetic energy of an electron.

$\bullet$ The effect of changing incident frequency v can also be studied from the plot of photoelectric current V_s potential applied across CD, keeping the intensity of incident radiation same.

From graph, we see that i_max is same in all cases.

(for same intensity).

From graph, as v increases, V_c becomes more negative.

$\bullet$ The plot of frequency Vs stopping potential for two different metals is shown here. It is clear from graph that there is a minimum frequency f₀ and f₀' for two metals for which the stopping potential is zero (V_c = 0). So for a frequency below f₀ and f₀' for two metals, there will not be any emission of photoelectrons. This minimum value of frequency of incident light below which the emission stops, however large the intensity of light may be, is called as threshold frequency.

Laws of Photoelectric Emission

1. For a light of any given frequency, photoelectric current is directly proportional to the intensity of light, provided the frequency is above the threshold frequency.

2. For a given material, there is a certain minimum (energy) frequency, called threshold frequency, below which the emission of photoelectrons stops completely, no matter how high is the intensity of incident light.

3. The maximum kinetic energy of the photoelectrons is found to increase with increase in the frequency of incident light, provided the frequency exceeds the threshold limit. The maximum kinetic energy is independent of the intensity of light.

4. The photo-emission is an instantaneous process. After the radiation strikes the metal surface, it just takes 10^–9 s for the ejection of photoelectrons.

SOURCE OF RADIATION

If P (in Watts) be the power of the source of radiation and I be the intensity of light radiation at a distance R from the source (R is the perpendicular distance of the receiving surface from the source), then:

$I=\frac{P}{4\pi R^2}$

If radiation is falling on a plate of area A, then, energy absorbed/sec by the plate is given as:

$Energy/\sec =I\times A=\left(\frac{P}{4\pi R^2}\times A\right)J/\sec$

If v be the frequency of the source (i.e. light radiation),

then energy per photon = hv

$\Rightarrow \text{number of photons falling on plate per second }=\frac{\text{energy per second}}{\text{energy per photon}}$

$\Rightarrow n/sec=\frac{\frac{P}{4\pi R^2}\times A}{hv}\left(Energy/photon=hv=\frac{hc}{\lambda }\right)$

If $v\ge v_0$ (v₀: threshold frequency of plate) and photon efficiency of plate be x %, then the number of photoelectron emitted per second is:

$\Rightarrow n/sec=\left(\frac{\frac{P}{4\pi R^2}\times A}{hv}\right)\times \frac{x}{100}$

HOW TO DETERMINE THE PHOTOELECTRIC CURRENT?

Let P be the power of a point source of electromagnetic radiations, then intensity I at distance r from the source is given by

$I=\frac{P}{4\pi r^2}\left(W/m^2\right)$

If A is the area of a metal surface on which radiations are incident, then the power received by the plate is

$P^{\prime }=IA=\frac{P}{4\pi r^2}A\left(W\right)$

If f is the frequency of radiation, then the energy of photon is given by

E = hf

The number of photons incident on the plate per second (called photon flux) is given by

$\Phi =\frac{P^{\prime }}{E}=\left[\frac{\frac{P}{4\pi r^2}\times A}{hf}\right]$

If $f>f_0$ (threshold frequency) and photon efficiency of the metal plate is $\eta$%, then the number of photoelectrons emitted per second is given by

$n=\frac{\Phi \eta }{100}\left[\frac{\frac{P}{4\pi r^2}\times A}{hf}\right]\frac{\eta }{100}$

Finally, the photocurrent i is given by

i = ne

Where e is the charge of an electron $\left(e=1.6\times {10}^{-19}J\right)$

Illustration : Photoelectric threshold of metallic silver is $\lambda$ = 3800 Å. Ultra-violet light of $\lambda$ = 2600 Å is incident on silver surface. Calculate

(i) the value of work function in joule and in eV.

(ii) maximum kinetic energy of the emitted photoelectrons.

(iii) the maximum velocity of the photo electrons.

(mass of the electron = 9.11 x 10^–31 kg)

Solution: (i) ${\lambda }_0=3800$ Å

$W=h{f}_0=h\frac{c}{{\lambda }_0}=\frac{6.633\times {10}^{-34}\times 3\times {10}^8}{3800\times {10}^{-10}}$

$J=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{3800\times {10}^{-10}}J=5.23\times {10}^{-19}J=3.27eV$

(ii) Incident wavelength $\lambda$ = 2600 Å

$\therefore f=incidentfrequency=\frac{3\times {10}^8}{2600\times {10}^{-10}}Hz$

Then $T_{\max }=hf-W_0$

$hf=\frac{6.63\times {10}^{34}\times 3\times {10}^8}{2600\times {10}^{-10}}=\frac{6.63\times 3}{2.6}\times {10}^{-19}J=7.65\times {10}^{-19}J=4.78eV$

$T_{\max }=hf-W_0=4.78eV-3.27eV=1.51eV.$

(iii) $T_{\max }=\frac{1}{2}m{v}_{\max }^2$

$\therefore v_{\max }=\sqrt{\frac{2T_{\max }}{m}}=\sqrt{\frac{2\times 2.242\times {10}^{-19}}{9.11\times {10}^{-31}}}=0.7289\times {10}^{6}m{s}^{-1}$