Physics Form 4 Notes : CHAPTER NINE – PHOTOELECTRIC EFFECT

CHAPTER NINE

PHOTOELECTRIC EFFECT

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 ultraviolet radiation, while caesium oxide needs visible light, i.e., optical spectrum (sunlight).

Work function

A minimum amount of work is needed to remove an electron from its energy level 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,

1 eV = 1 electron × 1 volt

1 eV = 1.6 × 10^-19 × 1 volt

1 eV = 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 possessed 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 have their energy quantified, i.e., it could only take discrete values, Max Planck predicted the energy of an oscillating atom to be

E = n h f, where n – integer, f – frequency of the source, h – Planck’s constant which has a value of 6.63 × 10^-34 Js.

Quantum theory of light

Planck published his quantum hypothesis in 1901, which assumes that the transfer of energy between light radiation and matter occurs in discrete units or packets. Einstein proposed that light is made up of packets of energy called photons which have no mass but have momentum and energy given by;

E = h f

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 ‘hf‘ must be greater than or equal to Φ. If ‘hf‘ 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.E_max = ½ m v²_max = h f – Φ ……………… (i), where m is mass and v_max is maximum velocity.

This is Einstein’s photoelectric equation.

If the photon energy is just equivalent to the work function, then m v²_max = 0. At this point, the electron will not be able to move, hence no photoelectric current, giving rise to a condition known as cut-off frequency, h f_co = Φ ……………… (ii)

Also, the potential difference required to stop the fastest photoelectron is the cut-off potential, V_co, which is given by E = e V_co electron volts. This energy is the maximum kinetic energy of the photoelectrons and therefore, ½ m v²_max = e V_co …………… (iii).

Combining equations (i), (ii), and (iii), we can write Einstein’s photoelectric equation as,

e V_co = h f – h f_co ………………….. (iv)

NOTE: Equations (i) and (iv) are quite useful in solving problems involving the 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’ = 3.0 × 10⁸ m/s. Since f = c / λ, then

   f = 3.0 × 10⁸ / 3.310 × 10^-7 = 9.06 × 10¹⁴ Hz.

2. The work function of tungsten is 4.52 eV. 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, h = 6.62 × 10^-34 Js).

   Solution:

   Frequency f = c / λ = 3.0 × 10⁸ / 2.50 × 10^-7.

   Energy of photon = h f = 6.62 × 10^-34 × (3.0 × 10⁸ / 2.50 × 10^-7) × (1 / 1.6 × 10^-19)

   = 4.97 eV.

   Hence hf_co = 4.52 eV. e V_co = 4.97 eV – 4.52 eV = 0.45 eV = 7.2 × 10^-20 J

   V_co = 7.2 × 10^-20 / 1.6 × 10^-19 = 0.45 V.

3. The threshold frequency for lithium is 5.5 × 10¹⁴ Hz. Calculate the work function for lithium. (Take h = 6.626 × 10^-34 Js)

   Solution:

   Threshold frequency, f_o = 5.5 × 10¹⁴ Hz, h = 6.626 × 10^-34 Js

   Φ = h f = 5.5 × 10¹⁴ × 6.626 × 10^-34 = 3.64 × 10^-19 J

4. Sodium has a work function of 2.0 eV. Calculate:
5. The maximum energy and velocity of the emitted electrons when sodium is illuminated by radiation of wavelength 150 nm.
6. Determine the least frequency of radiation by which electrons are emitted.

(Take h = 6.626 × 10^-34 Js, e = 1.6 × 10^-19 C, c = 3.0 × 10⁸ m/s, and mass of electron = 9.1 × 10^-31 kg).

Solution:

1. The energy of incident photon is given by h f = h c / λ

= (6.626 × 10^-34 × 3.0 × 10⁸) / 1.50 × 10^-7 = 1.325 × 10^-18 J

K.E_max = h f – 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 = ½ m v²_max. Therefore;

1.0 × 10^-18 = ½ × 9.1 × 10^-31 × v²_max

v²_max = (1.0 × 10^-18 / (0.5 × 9.1 × 10^-31)) = 2.2 × 10¹²

v_max = √(2.2 × 10¹²) = 1.5 × 10⁶ m/s (max. velocity of emitted electrons).

Φ = h f_co and f_o = Φ / h, Φ = 2 × 1.6 × 10^-19

f_o = (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 pressure). 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 soundtrack in a film.

Photoconductive cells – some semiconductors such as cadmium sulphide (CdS) reduce their resistance when light is shone on them (photoresistors). Other devices such as photodiodes and phototransistors block current when the intensity of light increases.

Photoconductive 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 meters.

2. 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 surface (this thin film allows light). The current increases with light intensity.

They are used in electronic calculators, solar panels, etc.