figure 1: Experimental demonstration of the photoelectric effect

## Introduction:

In the following article, I develop a simplified analysis of Einstein’s impressive 1905 paper on the photoelectric effect which demonstrates that radiation is quantised [1]. It is worth noting that this analysis was worth a Nobel prize in physics, and it was also very important in shedding light on Planck’s earlier discovery of a formula for black-body radiation that addressed the ‘ultraviolet catastrophe’.

While Planck encountered the notion of quantised electromagnetic radiation as a technicality while deriving his radiation law, he did not actually believe that electromagnetic radiation was quantised. Einstein, on the other hand, realised that the discrete description of electromagnetic radiation was fundamental. At a high level, he surmised that the continuous phenomena observed by Maxwell actually represented the average behaviour of discrete phenomena.

In an important sense, quantum theory developed initially as a hybrid theory in order to make sense of experiments which could not be explained using Maxwell’s electromagnetic theory.

## Description of the experiment:

In reference to the figure above, we make the following observations:

1. Knocking electrons free from the photoemission plate gives it a slight positive charge.

2. The photoelectric current generated by this method was quite small, but it could be measured with a micrometer.

3. The photoelectric current is also a proxy measure for the rate at which photoelectrons are leaving the surface of the photoemissive material.

4. When the power supply is set to a low voltage it traps the least energetic electrons, reducing the current through the micrometer.

5. Increasing the voltage drives increasingly energetic electrons back until none of them are able to leave the metal surface. The potential at which this occurs is the stopping potential. It turns out that this is a measure of the maximum kinetic energy of the electrons emitted as a result of the photoelectric effect.

6. Later experiments by Robert Millikan in 1914, showed that light with frequencies below a certain cutoff value, known as the threshold frequency would not eject photoelectrons from the metal surface no matter how bright the source was.

7. The result found by Millikan was quite unexpected since the energy in a beam of light is related to its intensity so classical physics would predict that a more intense beam of light(which has a greater number of photons) would eject electrons with greater energy than a less intense beam, no matter the frequency.

## Einstein’s insights:

Einstein and Millikan described the photoelectric effect using a formula that relates the maximum kinetic energy($$K_{max}$$) of the photoelectron to the frequency of the absorbed photons($$\upsilon$$) and the threshold frequency($$\upsilon_{0}$$) of the photoemissive surface:

$$K_{max} = h \cdot (\upsilon-\upsilon_{0})$$

Alternatively, we may represent $$K_{max}$$ in terms of the energy of the absorbed photons($$E$$) and the minimum work($$W_0$$) of the surface:

$$K_{max} = E-W_0$$

where we have:

$$E = h \cdot \upsilon = \frac{hc}{\lambda}$$

$$W_0 = h \cdot \upsilon_{0} = \frac{hc}{\lambda_{0}}$$

The maximum kinetic energy($$K_{max}$$) of the photoelectrons(with charge $$e$$) may be determined from the stopping potential($$V_0$$):

$$V_0 = \frac{W}{q} = \frac{K_{max}}{e} \implies K_{max} = e \cdot V_0$$

As for the rate at which photoelectrons are emitted from a photoemissive surface, $$\frac{\Delta n}{\Delta t}$$, this may be determined from the photoelectric current($$I$$):

$$I = \frac{\Delta q}{\Delta t} = \frac{\Delta n \cdot e}{\Delta t} \implies \frac{\Delta n}{\Delta t} = \frac{I}{e}$$

Finally, if we combine (2) and (5), we find that:

$$h \cdot \upsilon = W_0 + e \cdot V_0$$

which gives us the frequency as a function of $$V_0$$:

$$\upsilon(V_0) = \frac{e}{h} \cdot V_0 + \frac{W_0}{h}$$

## Conclusion:

Given that the above analysis concerns a single electron, for $$n$$ photons of light with frequency $$\upsilon$$ we should find that their intensity is proportional to their total energy which is given by:

$$E = n h \upsilon$$

Since Einstein’s discovery the photoelectric effect has been exploited in the development of night vision goggles and in photoelectron spectroscopy which is useful for determining the chemical composition of samples.

## References:

1. Albert Einstein. On a heuristic point of view about the creation and conversion of light. 1905.