Einstein’s Photoelectric Equation
Albert Einstein published an equation to explain this effect in 1905, the annus mirabilis (wonder year) of Physics. Light, according to Einstein, is a wave that interacts with matter as a packet of energy or a quantum of energy. The photon was the quantum of radiation, and the equation was known as Einstein’s photoelectric equation.
When a substance absorbs electromagnetic radiation, electrically charged particles are emitted from or inside it, resulting in the photoelectric effect. The phenomenon of emission of electrons by certain substance (metal), when it is exposed to radiations of suitable frequencies is called as photoelectric effect and emitted electrons are called photoelectrons.
The effect is frequently characterized as the ejection of electrons from a metal plate when exposed to light. The radiant energy may be infrared, visible, or ultraviolet light, X-rays, or gamma rays; the substance could be solid, liquid, or gas; and the particles discharged could be ions or electrons.
Characteristics of Photoelectric Effect:
- For each particular photosensitive material, there is a minimal cut-off frequency of the incoming radiation, known as the threshold frequency νo, below which no photo-electrons are produced.
- The threshold frequency varies depending on the metal.
- The photoelectric current is directly proportional to the intensity of incident light for a particular photosensitive material and the frequency of incident radiation (above threshold frequency).
- Above the threshold frequency ν0, the maximal kinetic energy of the emitted photo-electrons grows linearly with incident radiation frequency but is independent of incoming radiation intensity.
- The emission of photo-electrons occurs in an instant.
- There is no time lag between irradiation of the metal surface and photo-electronic emission.
Einstein`s Photoelectric Equation
In 1905, Einstein expanded Planck’s notion by deriving his own equation that correctly described the features of the photoelectric effect.
He presupposed two things.
- Radiation with frequency ν consists of a stream of discrete quanta or photons, with energy hν, where h is Planck’s constant. Photons travel at the speed of light through space.
- When photons and electrons in the emitter’s atoms collide when radiation of frequency ν is incident on a photosensitive surface. During such a collision, the photon’s whole energy is transmitted to the electron with no time lag.
A photon is not a material particle but quanta of energy. The incident photon’s absorbed energy hv by an electron is utilized in two ways. The electron uses some of its energy to break free from the atom. The minimum energy required to free-electron from a given surface is called photoelectric work function φ0 of the material of the surface.
The residual energy (h ν- φ0) emerges as electron kinetic energy. If the electron does not lose any of its energy in impact with the surface and exits with the greatest possible kinetic energy.
∴ 1 / 2 m v2max = hν – φ0 …..(1)
- m = mass of the electron
- vmax = maximum velocity of electron
All the photoelectrons emitted from the metal surfaces do not have the same energy.
The photoelectric current becomes zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy, so that the stopping potential (in volts) is numerically equal to the maximum kinetic energy of photo-electron in eV.
1 / 2 m v2max = ev0 …..(2)
- e = magnitude of the charge on an electron.
From (2) we have,
1 / 2 m v2max = hν – φ0
eV0 = hν – φ0
This equation is Einstein`s photoelectric equation.
With the aid of Einstein’s photoelectric equation, we can now describe all of the features of the photoelectric effect:
If the frequency of incident radiation is decreased, the kinetic energy of photoelectrons also decreases, and finally, it becomes zero for a particular frequency (say ν). ν is called the threshold frequency. Thus,
KEmax = 1/2 m V2max = 0
Therefore from equation (1), we get
0 = hν0 – φ0
hν0 = φ0
Therefore, Einstein`s equation can be written as,
1/2 m v2max = h(ν – ν0) …..(3)
From the above equation, we can say three points,
- ν > ν0, the photoelectrons are emitted with some velocity,
- ν < ν0, no photoelectrons are emitted, and
- ν = ν0, photoelectrons are just emitted with zero kinetic energy.
A more intense beam, according to quantum theory, includes a higher number of photons. As a result, the number of photon-electron collisions increases, and more photoelectrons are released. This explains why the photoelectric current increases with incoming photon intensity. Because the photoelectric work function (φ0) is constant for every given emitter, equation (3) demonstrates that the KEmax of photoelectrons grows with the frequency of incoming radiation but does not rely on intensity.
Photo-electrons are emitted as a result of electron-photon collisions. Such collisions occur as soon as the radiation strikes the photosensitive surface, and photoelectrons are released. There is no emission of photoelectrons at the incidence cut-off. As a result, the photoelectric effect occurs instantly.
Particle nature of light: Photon
The photoelectric effect, therefore, provided proof for the unusual fact that light acted as though it were formed of quanta or packets of energy, each of energy hν.
Is the light quantum of energy to be connected with a particle? Einstein discovered that the light quantum may also be related to momentum hv. A definite value of energy and momentum is a strong indication that the light quantum can be connected with a particle.
This particle was eventually given the name photon. The particle-like behaviour of light was established in 1924 by A.H. Compton’s (1892-1962) experiment on the scattering of X-rays from electrons. Einstein received the Nobel Prize in Physics in 1921 for his contributions to theoretical physics and the photoelectric effect. Millikan received the Nobel Prize in Physics in 1923 for his discoveries on the elementary charge of electricity and the photoelectric effect.
Summarized Photon picture of electromagnetic radiation
- When radiation interacts with matter, it behaves as if it were made up of particles known as photons.
- Each photon has energy E (= hu) and momentum p (= hν / c) where c is the speed of light.
- Whatever the intensity of radiation, all photons of light of a specific frequency ν or wavelength λ, have the identical energy E( = hν = hc / λ) and momentum p (= hν / c = h/λ). By raising the intensity of light of a certain wavelength, the number of photons per second crossing a given area increases, with each photon having the same energy. As a result, photon energy is independent of radiation intensity.
- Photons are electrically neutral and are unaffected by electric or magnetic forces.
- Total energy and total momentum are preserved in a photon-particle collision (such as a photon electron collision). However, the number of photons in a collision may not be preserved. The photon might be absorbed or a new photon could be produced.
Problem 1: An electron is accelerated from rest through a potential difference of 500 volts. Find the speed of the electron. (Given e = 1.6 × 10-19C, m = 19 ×10-31 Kg)
Voltage = 500 V
e = 1.6 × 10-19 C
m = 19 × 10 Kg
KE of emitted electron = eV = 1/2 m v2
= 2 × (1.6 × 10-19) × (500) / (9.1 × 10-31 )
V = √1.758 × 1014
= 1.326 × 107 m/s
Problem 2: A metal whose work function is 4.2 eV is irradiated by radiation whose wavelength is 2000 A°. Find the maximum kinetic energy of the emitted electron.
φ0 = 4.2eV = 6.72 × 10-19 J
λ = 2000 A°
1/2 m v2max = hν – φ0
KEmax = hν – φ0
= (hc / λ) – φ0
= (6.63 ×10-34 × 3 × 10) / (2 × 10-7) – (6.72 × 10-10)
= (9.945 – 6.72) × 10-19
= 2.015 eV
Problem 3: If photo-electrons are to be emitted from potassium surface with a speed of 6 ×105 m/s, what frequency of radiation must be used? The threshold frequency for potassium is 4.22 ×1014 Hz.
vmax = 6 × 105 m/s
ν0 = 4.22 × 1014 Hz
KEmax = 1/2 mv2max = h(ν – ν0)
ν = 1/2 × (mv2max / h) + ν0
= (1/2) × [(9.1 × 10-31 ×(6 × 105)2 / 6.63 × 10-34] +4.22 × 1014 Hz
= 2.47 × 1014 + 4.22 × 1014
= 6.69 × 1014 Hz
Problem 4: A photon of wavelength 3310A° falls on a photo-cathode and an electron of energy 3 × 10-19 J is ejected. if the wavelength of the incident photon is changed to 5000 A°, the energy of the ejected electron is 9.72 × 10-20 J. Calculate the value of Plank`s constant and threshold wavelength of the photon.
λ1 = 3310 = 3.31 × 10-7 m
KEmax (1) = 3 × 10-19 J
λ2 = 5000 = 5 × 10-7 m
KEmax = 9.72 × 10-20 J= 0.972 × 10-19 J
KEmax (1) = hc/λ1 – φ0
KEmax (2) = hc/λ2 – φ0
KEmax (1) – KEmax (2) = hc (1/λ1 – 1/λ2)
3 × 10-19 – 0.972 × 10-19 = h × 3 × 108 [(1/3.31 × 10-7) – (1/5 × 10-7)]
2.028 × 10-19 = h × 3 × 1015 (1.69 / 16.55)
h = 2.028 × 10-19 × 16.55 / 3 × 1015 × 1.69
= 6.62 × 10-34 Js
KEmax (1) = hc/λ1 – φ0
φ0 = hc/λ1 – KEmax (1)
= (6.62 × 10-34 × 3 × 108 / 3.31 × 10-7 ) – 3 × 10-19
= 3 × 10-19 J
φ0 = hc / λ0
λ0 = hc / φ0
= 6.62 × 10-34 × 3 × 108 /3 × 10-19
= 6.62 × 10-7
= 6620 A°
Problem 5: The Photo-electric function for a metal surface is 2.4 eV. If the light of wavelength 5000 A° is incident on the surface of the metal, find the threshold frequency and incident frequency, Will there be an emission of photo-electrons or not?
φ0 = 2.34eV = 3.84 × 10-19 J
λ = 5000A° = 5 × 10-7
ν = c / λ
= 3 × 108 / 5 × 10-7
= 6 × 1034 Hz
ν0 = φ0 / h
= 3.84 × 10-19 / 6.63 × 10-34
= 5.792 × 1014 Hz
As ν > ν0, photoelectric emission is possible.