This experiment was first done by Hertz in 1887. He discovered when light of high energy(frequency) was shone on metal, electrons were produced, later known as photoelectrons.

In his experiments, he found there exists threshold frequency $\nu_0$; such that when a light with a higher frequency $\nu > \nu_0$ is shone on the metal plate, a current is produced.

Also, the magnitude of the current is proportional to the light intensity, while the energy fo the photoelectrons emitted are independent of the intensity of light. (In later experiments, it was found that $E \propto \nu$ of the light)

In 1905, Einstein gave a feasible explanation: that light consisted of quanta (or photons, term coined by chemist Lewis), with energy $E=h\nu.$ If the amount of energy given to the electrons within the metal exceed a certain threshold $W$, electrons are released from the surface. The energy of the electrons coming out of the metal is equal to the energy of the photon minus the work function, i.e.

$E_{e^-}=\frac{1}{2}mv^2=h \nu-W$.

In 1915, Millikan verified Einstein’s conjecture through experimentation.

A UV light with $\lambda=290 \text{nm}$ is shone on a metal with $W=4.05 \text{eV}$. What is the energy $E_{e^-}$ of the released electrons, and what is their speed?

Note $\hbar = \frac{h}{2\pi}$, and $\hbar c \approx 200(197.33) \text{MeV}\times \text{fm}$, where $\text{fm}$ represents the fermi meter, $10^{-15}$ $\text{m}$.

$E_{photon}=h \nu=\frac{h c}{\lambda}=\frac{2\pi (197)\text{MeV} \times 10^{-15} \text{m}}{290*10^{-9}\text{m}}=4.28 \text{eV}$

Its speed is hence $\sqrt{\frac{2\times 4.28\text{eV}}{m_e}}=284\text{km/s}$.

Let’s first think about the units of $h$, Planck’s constant. If we denote $[x]$ as the units of $x$, and $M, L, T$ the units of mass, length and time:

$[h]=\frac{[E]}{[\nu]}=M(\frac{L}{T})^2/\frac{1}{T}=\frac{ML^2}{T}=M(\frac{L}{T})L=[p][r]$, which is the unit for angular momentum. This also motivates the equation of the de Broglie wavelength: $\lambda=\frac{h}{p}$. So now we have a relation between a particle and length! Well, how about when a particle is at rest? Doesn’t the denominator become zero and break the relation?

This begs the question: What is the wavelength of a particle with energy equal to its rest energy?

$E=mc^2=h\nu=\frac{hc}{\lambda}\rightarrow mc^2=\frac{hc}{\lambda}\rightarrow \lambda=\frac{h}{mc}$

This wavelength is called the Compton wavelength. In short, we just substitute $v=c$.

Fun fact: The Schwarzschild radius (the radius defining the event horizon of a black hole, a.k.a. the point of no return, $2GM/c^2$) and the Compton wavelength ($2 \pi \hbar /mc)$ corresponding to a given mass are similar when the mass is around one Planck mass ($m=\sqrt{\hbar G/c^3}$ ), when both are of the same order as the Planck’s length.

Protons are not only quanta for energy, but also for momentum.

$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} \text{ and } p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\rightarrow E^2-p^2c^2=m^2c^4$

So given the energy $E$ and momentum $p$, we can find the mass of any object. In non-relativistic scenarios, we have

$E=\frac{1}{2}mv^2 \text{ and }p=mv^2 \rightarrow E=\frac{p^2}{2m}$

$\text{For a photon, }m=0, \text{ so }E_\text{photon}=pc.\\\text{The momentum of a photon is thus }p=\frac{E_\text{photon}}{c}=\frac{h\nu}{c}=\frac{h}{\lambda}.$

Binding energies in atoms would be approx 10-13 eV.

In the experiment, we shine energetic X-rays on electrons. As a result,

This experiment has violated the classical Thomson Scattering. In this classical scattering, an EM wave is shown onto a free electron. The particle is assumed to undergo small amplitude oscillations about an equilibrium position that coincides with the origin of the coordinate system. Furthermore, the particle's velocity is assumed to remain sub-relativistic, which enables us to neglect the magnetic component of the Lorentz force. The electric field accelerates the electron which also periodically moves. When the electron is accelerated, the electron radiates photons. Technically, the energy is absorbed from the incident wave by the particle, and re-emitted as EM radiation.

The relationship between the radiation and the angle $\theta$ between the direction of acceleration of the particle and the direction of the outgoing radiation is

$\frac{d\sigma}{d\Omega}=(\frac{E}{mc^2})^2*\frac{1+\cos^2{\theta}}{2}$

For a full derivation of this equation, I highly recommend this website.

A property of this scattering is that the outgoing photons have the same frequency as the original wave. This implies the electron is being moved at the frequency of the electric field, and therefore the frequency of the radiation is the same.

However at high energies, Compton observed different results. When we treat the photon as a particle, our results should obey the conservation of energy and momentum. It can be shown that the photon cannot be absorbed by the electron by relativistic relations. The photon would also lose energy if it collides with the electron by transferring some of its energy to the kinetic energy of the electron. Energy and momentum conservation demand a shift in the frequency of scattered photons with respect to that of the incident photons. We should have $\lambda_\text{final}=\lambda_\text{initial}+k(1-\cos\theta)$. By dimensionality analysis, $k$ should be related to length. Therefore, $k$ must be related to the Compton wavelength $\frac{h}{m_e c}$ ,since the electron is the only existing particle in question.

The actual formula is $\lambda_\text{final}=\lambda_\text{initial}+\frac{h}{m_e c}(1-\cos\theta)$. At $\theta=0$, the photon just keeps going and somehow ignores the electron. For a photon travelling backwards with $\theta=\pi$, we have a factor of $2$ times the Compton shift. At $\theta=\frac{\pi}{2}$, we get the Compton shift.

$\lambda_C(e)=2.426\, \mathrm{pm}$

Q: A photon of wavelength $\lambda_i$ equal to the Compton wavelength of an electron $\lambda_C(e)$ hits an electron and is scattered exactly backwards. What is the velocity parameter $\beta=\frac{v}{c}$ of the recoiling electron?

A: Using Compton’s scattering formula, we have $\lambda_\text{final}=\lambda_\text{initial}+\frac{h}{m_e c}(1-\cos\theta)=\frac{h}{m_e c}+\frac{h}{m_e c}*2=\frac{3h}{m_e c}.$ By the conservation of momentum, the initial momentum is $m_ec$. The final momentum of the photon is $\frac{h}{\lambda_\text{final}}=\frac{1}{3}m_ec$ in the opposite direction. Therefore the final momentum of the electron is $\frac{4}{3}m_ec$. Since in this case the momentum is comparable to the rest energy, we have to treat the electron as relativistic. The relativistic momentum is

$p_ e =\frac{m_ e v}{\sqrt {1-\beta ^2}} \Rightarrow \frac{p_ e}{m_ e c} = \frac{\beta }{\sqrt {1-\beta ^2}}$(since $\beta =\frac{v}{c}$). We then find

$\frac{4}{3} = \frac{\beta }{\sqrt {1-\beta ^2}} \Rightarrow \beta = \frac{4}{5}.$

De Broglie mapped the particle of momentum $p$ to an associated wavelength $\lambda$ by the equation $\lambda=\frac{h}{p}$.