PHOTOELECTRIC EFFECT - Form 4 Physics Notes

Introduction

• Photoelectric effect was discovered by Heinrich Hertz in 1887.
• Photoelectric effect is a phenomenon in which electrons are emitted from the surface of a substance when certain electromagnetic radiation falls on it.
• Metal surfaces require ultra-violet radiation while caesium oxide needs a visible light i.e. optical spectrum (sunlight).

Work Function

• A minimum amount of work is needed to remove an electron from its energy level so as to overcome the forces binding it to the surface.
• This work is known as the work function with units of electron volts (eV ).
• One electron volt is the work done when one electron is transferred between points with a potential difference of one volt; that is,
1eV = 1 electron × 1 volt
1eV = 1.6 × 10-19 × 1 volt
1eV = 1.6× 10-19 Joules (J)

Threshold Frequency

• This is the minimum frequency of the radiation that will cause a photoelectric effect on a certain surface.
• The higher the work function, the higher the threshold frequency.

Factors Affecting the Photoelectric Effect

1. Intensity of the incident radiation – the rate of emission of photoelectrons is directly proportional to the intensity of incident radiation.
2. Work function of the surface – photoelectrons are emitted at different velocities with the maximum being processed by the ones at the surface.
3. Frequency of the incident radiation –the cut-off potential for each surface is directly proportional to the frequency of the incident radiation.

Planck’s Constant

• When a bunch of oscillating atoms and the energy of each oscillating atom is quantified i.e. it could only take discrete values.
• Max Planck’s
E = nhf , where n – integer, f – frequency of the source, h – Planck’s constant a value of 6.63 × 10-34 Js.

Quantum Theory of Light

• Planck’s published in 1901 his hypothesis which assumes that the transfer of energy in between light radiation and matter occurs in discrete quantum units or packets.
• Einstein proposed that light is made up of packets of energy called photons which have no mass but they have momentum and energy given by;
E = hf
• The number of photons per unit area of the cross-section of a beam of light is proportional to its intensity. However the energy of a photon is proportional to its frequency and not the intensity of the light.

Einstein’s Photoelectric Equation

• As an electron escapes energy equivalent to the work function ‘Φ’ of the emitter substance is given up. So the photon energy must be greater be greater than or equal to Φ.
• If h is greater is greater than Φ then the electron acquires some kinetic energy after leaving the surface.
• The maximum kinetic energy of the ejected photoelectron is given by;
K.Emax = ½ mv2max = hf – Φ ………………(i), where mv2max = maximum velocity and mass. This is the Einstein’s photoelectric equation
• If the photon energy is just equivalent to work function then, mv2max = 0, at this juncture the electron will not be able to move hence no photoelectric current, giving rise to a condition known as cut-off frequency , hfco = Φ………………. (ii)
• Also the p.d required to stop the fastest photoelectron is the cut-off potential , Vco which is given by E = eVco electron volts, but this energy is the maximum kinetic energy of the photoelectrons and therefore,
½ mv2max = eVco ………….. (iii)
• Combining equations (i), (ii) and (iii), we can e V co = hf – hfco ………………….(iv)
NOTE: -- Equations (i) and (iv) are quite useful in solving problems involving photoelectric effect.

Examples
1. The cut-off wavelength for a certain material is 3.310 × 10-7 m. What is the cut-off frequency for the material?

Solution
Speed of light, c = 3 × 108 m/s.0.
Since  f = c/λ then f = 3.0 × 108/3.310 × 10-7 = 9.06 × 10 14 Hz.
2. The work function of tungsten is 4.52eV. Find the cut-off potential for photoelectrons when a tungsten surface is illuminated with radiation of wavelength 2.50 × 10-7 m. (Planck’s constant = 6.62 × 10-34 Js).

Solution
Frequency f = c/λ /  = 3× 10/2.50 × 10-7
Energy of photon = hf = 6.62 × 10-34 × (3.0 × 108/2.50 × 10-7 ) × (1/1.6 × 10-19) = 4.97 eV.
Hence hfco = 4.52 eV.
eVco = 4.97 eV - 4.52 eV = 0.45 eV = 7.2 × 10-20 J
Vco = 7.2 × 10-20/1.6 × 10-19 = 0.45 eV.
3. The threshold frequency for lithium is 5.5 × 1014 Hz. Calculate the work function for lithium. (Take  ‘h’ = 6.62 × 10-34 Js)

Solution
‘h’ 6.626 × 10-34 Js
Threshold frequency, fo = 5.5 × 1014 Hz
Φ = hf = 6.626 × 10-34 × 5.5 × 1014 = 36.4 × 10-20
4. Sodium has a work function of 2.0 e V. Calculate
1. The maximum energy and velocity of the emitted electrons when sodium is Illuminated by a radiation of wavelength 150 nm.
2. Determine the least frequency of radiation by which electrons are emitted. (Take ‘h’ = 6.626 × 10 Js,  e = 1.6 × 10-19, c = 3.0 × 108 m/s and mass of electron = 9.1×10-31 kg).

Solution
a) The energy of incident photon is given by hf
= c/λ = (6.626 × 10-34 × 3.0 × 108)/1.50 × 10-9 = 1.325 × 10-18 J
K.E max = hf – Φ = (1.325× 10-18) – (2 × 1.6 × 10-19) = 1.0 × 10-18 J (max. K.E of the emitted electrons)
But K.E max = ½mv2max .
Therefore; 1.0 × 10-18 = ½ × 9.1 × 10-31 × V2max
V2max = (1.0 × 10-18/9.1 × 10-31) 1/2 = 1.5 × 106 m/s (max. velocity of emitted electrons).
b) Φ = fco and fo = Φ/λ, Φ = 2 × 1.6 × 10-19
fo = (2 × 1.6 × 10-19)/(6.626 × 10-34) = 4.8 × 1014 Hz (min. threshold frequency of the emitted electrons)

Applications of Photoelectric Effect

1. Photo-emissive cells
• they are made up of two electrodes enclosed in a glass bulb (evacuated or containing inert gas at low temperature).
• The cathode is a curved metal plate while the anode is normally a single metal rod)
• They are used mostly in controlling lifts (doors) and reproducing the sound track in a film.
2. Photo - conductive cells
• Some semi-conductors such as cadmium sulphide (CdS) reduces their resistance when light is shone at them (photo resistors).
• Other devices such as photo-diodes and photo-transistors block current when the intensity of light increases.
• Photo-conductive cells are also known as light dependent resistors (LDR) and are used in alarm circuits i.e. fire alarms, and also in cameras as exposure metres.
3. Photo - voltaic cell
• This cell generates an e.m.f using light and consists of a copper disc oxidized on one surface and a very thin film of gold is deposited over the exposed surfaces (this thin allows light).
• The curent increases with light intensity.
• They are used in electronic calculators, solar panels etc.

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