In 1905, the annus mirabilis (miracle year) of Physics, Albert Einstein proposed an equation to explain this effect. Einstein argued that light was a wave that interacts with matter in the form of a packet of energy or a quantum of energy. This quantum of radiation was a photon and the equation was called Einstein’s photoelectric equation.

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## Einstein’s Photoelectric Equation

### The Photoelectric Effect

Let us recall the Photoelectric effect.

- Firstly, above a specific value of the frequency (threshold frequency), the strength of the photoelectric current depends on the intensity of the light radiation.
- The reverse potential at which the photo-current stops (stopping potential) is independent of the intensity of light. Therefore, no matter how intense your source of light is, it can’t defeat the stopping voltage.
- Any values of frequency below the threshold value are unable to produce a photoelectric current. Therefore, even if you take a metallic strip to the surface of your nearest star (Sun), you will never get a photocurrent if the frequency of the radiation is smaller than the threshold frequency.
- The photoelectric effect was almost instantaneous. This meant that as soon as you turn your source of light on, pop goes the electron!

### Enter Einstein and His Equation of The Photoelectric Effect

Einstein’s view of light was magnificent as well as revolutionary. He proposed a weird but effective model of radiation. Light consisted of very small particles. These particles were not matter but pure energy. He called each of these a quantum of radiation. Therefore, light must be made up of these quantas or packets of energy or quantum energy. We call them photons and they carry the momentum and energy from our source of light.

According to the Einstein-Plank relation, we have E = hν …(1)

Where ‘h’ is the Plank’s constant and ‘ν’ is the frequency of the radiation emitted.

Also from the experiment on Photoelectric effect, we see that there is a threshold frequency below which the electrons won’t come out of the metallic surface. In equation (1) we see that Energy is a function of frequency. Hence this observation is explained by equation (1). This also explains the instantaneous nature of the photoelectric emission.

Once a photo-electron is outside the metallic surface, what will be its energy? Since there is no electric field outside the metal surface, the energy of an electron will be purely Kinetic in nature. The quantum energy absorbed from the photon will be partly used to overcome the attraction of the metallic surface.

So, we have K.E. of the photo-electrons = (Energy obtained from the Photon) – (The energy used to escape the metallic surface)

This energy is a constant for a given surface. We denote it by Φ. We call it the work function and it is constant for a given substance. Thus we can write:

K.E. = hν – Φ … (2)

This is the Einstein’s Photoelectric equation.

#### More About Work Function

Imagine a football trapped inside a tub. Let us say that you have to get this football out of the tub by hitting it with smaller balls. The balls you throw in the tub must have a minimum energy in order to be able to extract the football from the tub. This energy is the work function of the tub and the football.

Similarly, an electron needs some minimum energy to be extracted from a metallic surface. In equation (2), if ν = threshold frequency (ν_{0}) then the electrons will have just enough quantum energy to come out of the metal. The Kinetic Energy of such an electron will be supposedly zero. Since it only gets energy enough to liberate itself from the metal surface. using these values of ν and K.E. in equation (2), we have:

hν_{0} – Φ = 0 or hν_{0 }= Φ ….(3)

using in (2), we have K.E. = hν – hν_{0}

or K.E. = h(ν – ν_{0})

Also if V_{0} is the Stopping Potential, then

K.E. _{(max)} = eV_{0}; using this in equation (3), we have:

eV_{0} = h(ν – ν_{0}) ……(4)

The values of ‘h’ are got from the photoelectric experiment by the above equation. The values so obtained were in agreement with the actual values and thus confirmed Einstein’s explanation of the Photoelectric effect.

## Solved Examples For You

**Example 1**: If a photocell is illuminated with a radiation of 1240A^{o}, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength are

A) 1eV, 5200 A^{0} B) 2eV, 6200 A^{0}

C) 3 eV, 7200 A^{0} D) 4eV, 4200 A^{0}

Solution: B) We know that ν = c/λ. Thus Einstein’s equation for Photoelectric effect can be written as K.E. _{(max)} = hc/λ – Φ …(5)

K.E. _{(max)} = eV_{0} = e (8V) ; using this in (5), we have:

8e = hc/1240×10^{-10} – Φ

Substituting h = 6.62×10^{-34} J.s and c = 3×10^{-8} m/s, we get

Φ = 3.2×10^{-19} J or 2 eV (1eV = 1.6×10^{-19} J)

Also, we have hν_{0 }= Φ or hc/λ_{0 }= Φ and λ_{0} = 6200 A^{0}