# Photoelectric effect and quantum physics

When metals absorb electromagnetic radiation above a certain frequency, they emit electrons called photoelectrons. This phenomenon is called the photoelectric effect.

The **minimum frequency** of the incident radiation that can induce the photoelectric effect is called the threshold frequency $$(f_{0})$$. It varies from metal to metal.

If the frequency of the radiation is below the threshold, it cannot trigger the photoelectric effect. Even increasing the intensity (the number of photons hitting the metal) does not change that.

Albert Einstein found that the existence of a threshold frequency could only be explained if **electromagnetic radiation is quantised** (i.e. it comes in discrete packets: photons). He said that **each emission of a photoelectron is triggered by only a single photon**. Increasing the number of photons therefore does not

Because each photoelectron emission is triggered by only one photon, photons with too low a frequency (and hence energy) are unable to cause photoelectron emission regardless of how many photons hits the metal surface.

Einstein's photoelectric equation describes the kinetic energy of a photoelectron emitted from a metal surface through the photoelectric effect.

The equation only gives **the maximum kinetic energy attainable** ($$K_\text{max}$$) by a photoelectron.

This is because the energy that the photon transfers to the electron upon hitting the atom is random but cannot exceed the photon's total energy before the collision.

The maximum kinetic energy is equal to the energy of the photon triggering the photoelectric effect $$hf$$ and the work function $$\phi$$ (the energy needed for the photoelectric effect to be triggered on a specific metal): $$$K_\text{max}=\frac{1}{2}mv_{\text{max}}^{2}=hf-\phi$$$

The maximum kinetic energy of photoelectrons is **independent of the intensity of the incident radiation** (it only depends on frequency).

The frequency of visible light is below the threshold frequency for most materials and hence it will not trigger the photoelectric effect in most cases.

$$m=$$ mass of an electron (a constant); $$v_{\text{max}}=$$ maximum velocity of the electron; $$h=$$ Planck constant; $$f=$$ frequency of the incident radiation; $$hf$$ energy of an incident photon; $$K_{\text{max}}=$$ maximum kinetic energy of a photoelectron; $$\phi=$$ work function of the metal.

The **minimum energy** a photon must have to trigger the photoelectric effect in a metal is called the work function $$(\phi)$$ of the metal.

The work function is minimum amount of energy that needs to be transferred to an electron for the electron to be emitted. This energy is equivalent to the energy of a photon that has a frequency equivalent to the threshold frequency $$f_{0}$$ of the metal: $$$\phi=hf_{0}$$$

If an incident photon has energy equivalent to the work function and transfers all of its energy to an electron, the emitted electron will have no kinetic energy left once it has left the metal surface.

This is similar to the minimum kinetic energy a spaceship requires to escape the gravitational field of a planet.

The threshold frequency needed for a photon to trigger the photoelectric effect is related to the **escape velocity** of an object trying to leave a planet.

The difference is that the threshold frequency describes a **requirement for the photon** triggering the emission, not for the electron that is actually leaving the atom.

$$f_0=$$threshold frequency; $$h=$$ Planck constant; $$\phi=$$work function.

The photoelectric effect causes electrons to be emitted at a certain rate, resulting in an **electric current** between the surface of the metal and a point outside of the metal. This current is known as the photoelectric current.

A metal surface designed to emit photoelectrons for the purposes of generating a photoelectric current is called an **emitter**. A metal surface designed to collect photoelectrons from the emitter is called a **collector**.

Formally, the photoelectric current $$I$$ is the rate of flow of charge (in the form of photoelectrons) from an emitter to a collector: $$$I=\frac{ne}{t}$$$ $$n$$ is the number of photoelectrons collected within a time $$t$$; $$e$$ is the charge of an electron ($$e=1.6\times 10^{-19}\text{ C}$$).

**Intensity** is the total power (i.e. energy/time) of a wave per unit area normal to the direction of movement of the wave.

For an electromagnetic wave, the intensity is $$$ \begin{align*} \text{Intensity}&=\frac{\text{Total energy of incident photons}/\text{Time}}{\text{Normal area}}\\ &=\frac{P}{A}=\frac{N(hf)/t}{A} \end{align*} $$$ $$N$$ is the number of incident photons, $$hf$$ is the energy of a photon, $$t$$ is the time period during which N photons hit a surface and $$A$$ is the area normal (i.e. perpendicular) to the direction of the wave motion.

In the photoelectric effect, the number of photoelectrons emitted $$(n)$$ and the number of incident photons $$(N)$$ are **directly proportional** ($$n\propto N$$).

$$n$$ is not equal to $$N$$ because the amount of energy transferred from photons to electrons varies randomly and some electrons will not get enough energy to escape from the atom.

The proportionality between photons and photoelectrons implies that the photoelectric current and the intensity of incident radiation are also directly proportional: $$$n\propto N \quad \Rightarrow \quad I\propto \text{Intensity}$$$

These proportionalities require that the frequency of the incident radiation be **greater than the threshold frequency** (i.e. $$f\gt f_{0}$$). There would be no photoelectrons if the frequency is lower than the threshold frequency.

When a collector is at a **lower potential** than an emitter, the electrons travelling from the emitter to the collector **lose kinetic energy** and **gain** an equivalent amount of **electric potential energy**.

This is because the electric field between the two plates is directed from the emitter to the collector.

The electric force (recall $$\vecphy{F}=q\vecphy{E}$$) on the negatively charged photoelectrons points in the opposite direction of the electric field, causing them to decelerate.

The photoelectrons therefore lose kinetic energy by moving in the direction of the electric field.

The stopping potential $$(V_{\text{s}})$$ is the potential difference required to stop a photoelectron with the maximum possible kinetic energy exactly at the collector's surface (i.e. the velocity becomes zero when it reaches the surface).

This implies that **all of the photoelectron's kinetic energy is converted into potential energy**.

**Einstein's photoelectric equation** can then be rewritten as: $$$ \begin{align*} eV_{\text{s}}&=\frac{1}{2}mv_{\text{max}}^{2}=hf-\phi\\ \Rightarrow V_{\text{s}}&=\left (\frac{h}{e}\right )f-\frac{\phi}{e} \end{align*} $$$ $$e$$ is the electron charge (a constant), $$m$$ is the electron mass, $$v_{\text{max}}$$ is the maximum velocity of the electron, $$f$$ is the frequency of the incident radiation and $$\phi$$ is the work function of the metal.

A graph of $$V_{\text{s}}$$ vs $$f$$ would give a **constant gradient** of $$h/e$$ with a y-intercept of $$-\phi/e$$.