Quantum Physics at A level and beyond
(For the OCR A syllabus and beyond)
The photon is a light particle, discovered by Max Planck. It came about because electromagnetic energy could only take certain discrete values and cannot be a continuous wave due to the Ultraviolet Catastrophe. A photon is thus a quantum of light.
The energy of the photon can be calculated by the following equation, which shows that the energy of a photon can be determined through its frequency and is directly proportional to this. The constant h is known as Planck’s constant which is 6.62607004 × 10–34m2 kg / s.
Recall the wave speed equation:
Rearrange as such:
Then substitute the resulting f back into the original equation:
This gives you the energy if you know the wavelength.
Units of Quantum Mechanics:
Basically everything is small, especially including the amount of energy at a quantum level. A joule is huge and a typical gamma photon (the most energetic kind) have 3*10^-19J of energy. So we came up with a whole new system of units to describe them because that’s just easier. One electron with an energy of 1eV has that energy because it moved through a 1V potential difference. The work done on a charged particle moving through a potential difference is the charge on the particle multiplied by the potential difference.
Where 1.60*10^-19C represents the charge in Coulombs.
LEDs and the Planck constant :
The Planck constant can be found with the use of a simple LED experiment.
An LED starts emitting photons in the visible light section of the electromagnetic spectrum when the threshold p.d. is reached. The work done at this potential difference is given by this threshold p.d. multiplied by the charge Q, or W = E = Qv.
This work done is approximately the same as the energy of the photon, though there will be differences depending on efficiency and measure. So therefore E = Qv ≈ hf. Recall the rearranged equation to find energy from wavelength where h and c (the speed of light) are constants:
Therefore we can equalise the original E = QV equation:
The charge on an electron can be represented with e:
Since e, h and c are all constants with h being the constant to find, we can show that the voltage is inversely proportional to λ, or the voltage is proportional to 1/λ.
We already know the wavelengths of each LED, so we just need to find out the threshold voltage of each LED.
To find the Planck constant we can plot a linear graph that passes through the origin with threshold p.d on the y-axis and 1/λ on the x-axis. The gradient of this would be hc/e and dividing the gradient by the constant c/e or speed of light over charge on an electron, gives you the Planck constant.
An algebraic method can be through this:
A graphical method would look like this (shamelessly copypasted from the textbook):
To increase the accuracy of our result, we can do several tests with different wavelength LEDs, then take an average of all the values of h found.
The photoelectric effect:
When electromagnetic radiation hits a metal, electrons are launched off the surface. This happens because photons excite the electrons to the point that they have so much energy that they are able to escape the atom, this is photoelectric ionisation. As well as this, the photoelectric effect only occurs at discrete amounts of electromagnetic energy. This is because the atom has discrete energy levels. The electrons emitted are referred to as photoelectrons because they originated through the photoelectric effect.
The gold-leaf electroscope and charging:
If you touch the top plate with a high voltage power supply, the gold-leaf electroscope becomes charged. This means that there are negative charges everywhere and this can demonstrate the repulsion between like charges.
You can use a gold leaf electroscope to demonstrate the photoelectric effect. If you put a piece of zinc on top of the disc and shine high energy radiation (such as Ultraviolet rays) on it, the gold leaf will become uncharged and coil back into itself. This is because due to the photoelectric effect, the electrons on the gold leaf were ejected from the surface leading to the gold leaf becoming uncharged.
More about the photoelectric effect:
Note: a higher frequency means more energy in a photon. A greater intensity refers to more photons per unit time.
Electrons are only emitted from the metal surface if the photon frequency is greater than a particular amount, this is called the threshold frequency. Intensity or energy density doesn’t matter in this case.
If the frequency of the incident radiation is above the threshold, electrons would be emitted instantly.
Finally, because increasing the intensity is really just the same thing as emitting more incident photons, the intensity does not affect the kinetic energy of the emitted photoelectrons. Instead, it leads to the emission of more photoelectrons. So therefore more photons mean more electron excitations, therefore more electrons are emitted.
The distinction made at the beginning between intensity and energy via frequency is the reason that the regular wave model cannot explain the photoelectric effect, yet the photon model can since it models electromagnetic radiation as a stream of Photon particles.
His suggestion was that each electron needs only a specific amount of energy in order to escape, and a photon interacting with the electron can transfer its energy through absorption and followed excitation of the electron. Because the energy of a single photon is actually solely dependent on its frequency, a greater number of photons with a smaller-than-threshold frequency wouldn’t really make a difference at all. All a greater intensity is, is more photons per unit of time which leads to more photoelectrons emitted. The rate of electron emission depends on the intensity or rate of incident photon absorption.
A greater photon intensity makes no difference if the photon doesn’t have the threshold frequency.
A greater photon frequency means a greater kinetic energy electron emitted.
A greater photon intensity means a greater number of electrons emitted.
Einstein also explained why a threshold frequency was required and different for different metals. This was due to the position of the electron within the metal lattice. If it was deeper within the lattice then it required more energy to be released. He defined the energy required to escape as the work function.
He used this to create the photoelectric effect equation, in combination with the principle of conservation of energy.
Einstein’s photoelectric effect equation:
First, we must note that if the energy of an incident photon is above the threshold value, the remaining energy (photon energy — threshold value) is converted to the kinetic energy of the electron emitted. This is due to the principle of conservation of energy. From this we can derive:
Where the energy of the photon (hf) = the work function (Φ) represents the work function or minimum amount of energy required to displace the electron. KEmax represents the maximum kinetic energy of an emitted photoelectron due to this.
All units need to remain constant throughout. Energies should be in an order of magnitude of electronvolts (eV) or Joules (J).
We can demonstrate this equation and the threshold frequency through a graph (again, shamelessly copypasted from the textbook)
In this graph, f0 represents the threshold frequency, the minimum frequency required for electron excitation.
Up until x=f0, no electrons are emitted because the photon frequency is simply too low. From f0 onwards, the maximum kinetic energy is proportional to f-f0, as demonstrated by Einstein’s equation. The gradient is then = h (because E = hf).
Different metals have different work functions depending on where the electron is in the lattice.
So very small, quantum realm particles can act as waves. According to de Broglie, all matter has wave-like properties and particle-like properties.
We can’t observe the wave-like properties of an electron unless we diffract it through a diffraction grating with gaps similar to the wavelength of the electron itself.
This is described through the dual slit experiment, in which electrons are fired through a diffraction grating showing an interference/diffraction pattern on the screen.
This is also the case for photons, they are seen as particles and yet they diffract.
The de Broglie equation relates a wave-acting particle’s wavelength to its momentum:
This was de Broglie’s first observation, that the wavelength is inversely proportional to momentum.
This eventually led to the de Broglie equation:
This uses the Planck constant to relate the wavelength and momentum.
Crystallography is a method for determining molecular structure. It was used in the development of the dual helix model of DNA. X-ray crystallography involves firing X-rays which are diffracted, and the resulting diffraction patterns can aid us in finding molecular structures. Electron crystallography can also be used for this, but electrons can actually be better. Their wavelength can be tuned, for example by tuning the kinetic energy through the de Broglie relation. They can also be used with very thin sheets or layers of materials.
[NOT A LEVEL] The Ultraviolet Catastrophe and the development of the photon:
The Ultraviolet Catastrophe provided the basis for Max Planck’s proof of the Photon. It began with radiation emitted from materials at high temperatures, or black-body radiation. If there was a single continuous wave, then at approaching higher frequencies, such as the ultraviolet region, the current classical, thermodynamic view of black-body radiation predicted an infinite intensity (or amplitude) at these higher frequencies. This is physically impossible so Planck decided to attempt to “chop up” the energy into discrete arbitrary chunks which fixed this problem as these “particles” could no longer lead to an infinite intensity. This lead to the photon; a packet of discrete electromagnetic energy.
[NOT A LEVEL] Electromagnetic wave model and revision:
There are two major laws of Electrodynamics (the Physics of changing electromagnetic fields). One states that a changing magnetic field creates an electrical field and one states that a changing electrical field creates a magnetic field. Irrespective of the source, because it is the fields themselves that are moving. This is the basis of electromagnetic waves, a moving field is produced and this induces more fields that move and propagate through space. Consider a radio transmission tower that produces radio waves, it has electrons that move up and down. Now consider the wave created by this tower, there is a delay between the origin of the wave and a node of the wave chosen at any other point. From this delay and the wavelength, we can find the speed; a constant which is 3*10⁸ and remains the same regardless of wavelength or frequency (v=fλ). Scientists believed earlier that electromagnetic radiation was one continuous wave, the belief of classical physics. Due to the ultraviolet catastrophe, this view was revised by Planck.
[NOT A LEVEL] Uses and alternatives of the photoelectric effect:
Solar panels work using something similar to the photoelectric effect known as the photovoltaic effect. In semiconductor physics, the amount of energy an electron has can lead to it either staying in the shell of its atom or getting conducted. An electron moving in this way is called moving from the valence band to the conduction band. So, the energy from UV radiation from the sun is transferred to the electron raising it to the conduction band and this leads to the electron moving, inducing a potential difference. Another thing you must note is that it isn’t ionised because the electrons flow in a circuit so lost electrons are replaced.
Image sensors and cameras actually work in a very similar way, such as CCDs or charge-coupled devices. In these, however, charge accumulates and is measured afterwards to form an image.
Spacecraft can be affected negatively by the photoelectric effect, because it can get ionised by the incident radiation causing a positively charged exterior. This can cause the failure of electronic equipment.
[NOT A LEVEL] Derivation of the de Broglie relation:
E = mc²
E = mc⋅c
E = p⋅c (momentum(p)= mc)
hf = p⋅c (E = hf)
h = p⋅λ
h/p = λ
[NOT A LEVEL] Photons have momentum:
The de Broglie relation can also be used to define photons, as they exhibit WP duality. But, the de Broglie relation uses momentum p. p is defined as p=mv but we know that photons have no mass. Although photons don’t have mass, they still have momentum and this is because of mass-energy equivalence. E = mc², or the lesser-known E² = m²c⁴+p²c² states that mass and energy are essentially the same thing but in a different way, so we can say that because the photon has energy, it has relativistic mass, which means that it can have momentum.
The idea that photons have momentum is especially useful in Compton scattering and to make ultracold atoms using lasers. The way we can do this is by shooting lasers at an atom in all directions. The photons carry momentum and therefore the atom stays completely still causing it to cool down to just billionths over absolute zero.
[NOT A LEVEL] Compton scattering:
When a photon hits an electron, the electron and the photon both scatter. This can be modelled by an elastic collision in two dimensions (Further Mechanics) and we know that momentum is conserved. However, the scattered photon has a greater wavelength in comparison to the original photon. This makes sense from a quantum perspective. Energy needs to be conserved and a single photon has a set intensity and therefore the frequency of the scattered photon will decrease due to the relation E=hf. If we know the difference in wavelength, we can find out the energy carried by the electron and therefore the velocity of the electron with 1/2mv².
[NOT A LEVEL] Light intensity (and everything else in quantum mechanics for that matter) is probabilistic:
From a classical perspective, the intensity of light is given by the amplitude of the wave, however, from a quantum perspective, the intensity of light is given by the relative probability of a photon crossing a barrier. For example the intensity of light coming out of a polarising filter is given by the probability that a photon will cross the filter. This can actually be given using vector multiplication in which we multiply the vector that defines the photons original polarisation with the vector that defines the filter’s polarisation. This should give the probability that a photon goes through and therefore the intensity.