12.1 The interaction of matter with radiation

The photoelectric effect

Photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.

Photoelectric Effect Demonstration

Light consists of photons of energy hf. Each photon interacts with a single electron. There is a minimum threshold frequency below which no electron is emitted. Energy is needed to overcome the attractive forces on the electron within the metal (work function). Any further supplied energy becomes kinetic energy of the electron. Increasing the intensity of light increases the emission rate of the electrons.

\begin{gather} E_{max} =hf-\Phi \\ \notag\\ E_{max} =maximum\ kinetic\ energy\ of\ emitted\ electron \notag\\ \Phi =work\ function,\ hf=energy\ of\ the\ photon\ \notag\\ ( f\ is\ frequency\ of\ light,\ h\ is\ planck\ constant) \notag \end{gather}

Wave-particle duality gives an explanation for the photoelectric effect.

Video explanation
Video explanation #2

Matter waves

De Broglie wavelenght:

\begin{gather} \lambda =\frac{h}{p}\\ \lambda =wavelenght\ of\ the\ particle\ ( de\ Broglie\ wavelenght) \notag\\ h=planck's\ constant,\ p=momentum\ of\ particle\ ( mv) \notag \end{gather}

Developed by French physicist Louis de Broglie in 1924.

Video explanation

Electron diffraction

Electron diffraction refers to the wave nature of electrons.

The Davisson–Germer experiment was a 1923-27 experiment by Clinton Davisson and Lester Germer at Western Electric (later Bell Labs), in which electrons, scattered by the surface of a crystal of nickel metal, displayed a diffraction pattern.

\begin{gather} \lambda =\frac{h}{\sqrt{2meV}}\\ \notag\\ \lambda =de\ Broglie\ wavelenght,\ h=Planck's\ constant \notag\\ m=mass\ of\ electron,\ e=charge\ of\ electron,\ V=voltage \notag \end{gather}
Video demonstration
video explanation

The Bohr model

Bohr model assumptions:

1. Electrons in atom exist in stationary states

2. Electrons move between stationary states by absorbing or emitting a quantum of EM radiation

3. The angular momentum of electron in a stationary state is quantized in integral values of ʰ⁄₂π

* This model cannot explain more complicated systems of atoms and is a big simplification.

Energies in the Bohr orbits

For hydrogen atom:

\begin{gather} E=-\frac{13.6}{n^{2}}\\ \notag\\ E=energy\ in\ orbit,\ n=energy\ level\ ( principal\ quantum\ number) \notag \end{gather}
Video explanation
Video explanation #2

Schrödinger's equation

Wave-particle duality explains a bright interference fringe as being the place with a high probability of finding a particle. The position of particles is mathematically described by probability waves.

Schrödinger's wave function ψ describes the quantum state of particles.

When the square of the amplitude is a maximum, there is the greatest probability of finding a particle. When the square of the amplitude is 0, there is 0 probability of finding a particle.

The square of the wave function is proportional to the probability per unit volume of finding a particle (probability density)

\begin{gather} P( r) =|\psi |^{2} \Delta V\\ \notag\\ P( r) =probability\ of\ finding\ particle\ at\ distance\ r\ from\ origin \notag\\ \Delta V=volume\ concidered,\ \psi =wave\ function \notag \end{gather}

Copenhagen interpretation - nothing is real unless it is observed.

Video explanation
Copenhagen interpretation

The Heisenberg uncertainty principle

The uncertainty principle states that the more precisely the position of some particle is determined, the less precisely its momentum can be predicted from initial conditions, and vice versa.

\begin{gather} \Delta x\Delta p\geqq \frac{h}{4\pi }\\ \Delta x=uncertainty\ of\ position,\ \Delta p=uncertainty\ of\ momentum \notag\\ h=Planck's\ constant \notag \end{gather}
Heisenberg's Uncertainty Principle Explained
Uncertainty principle video (optional but highly recommended)

Pair production and annihilation

Pair production, in physics, formation or materialization of two electrons, one negative and the other positive (positron), from a pulse of electromagnetic energy traveling through matter, usually in the vicinity of an atomic nucleus. Pair production is a direct conversion of radiant energy to matter. It is one of the principal ways in which high-energy gamma rays are absorbed in matter. For pair production to occur, the electromagnetic energy, in a discrete quantity called a photon, must be at least equivalent to the mass of two electrons.

Minimum energy needed for pair production:

\begin{gather} E=2mc^{2}\\ \notag\\ m=rest\ mass\ of\ particle,\ c=speed\ of\ light\ in\ vacuum,\ \notag\\ E=energy \notag \end{gather}

Electron-positron pair production

When a particle meets its antiparticle they annihilate, forming two photons of energy equal to the total mass-energy of the annihilating particles.

Bubble chamber tracks of electron-positron pair production.

Pair production and uncertainty principle

Energy and time are also conjugate variables.

\begin{gather} \Delta E\Delta t\geqq \frac{h}{4\pi }\\ \notag\\ \Delta E=uncertainty\ of\ energy,\ \Delta t=uncertainty\ of\ time \notag\\ h=Planck's\ constant \notag \end{gather}
Video explanation
Hawking radiation (optional)

Quantum tunneling

Quantum tunneling is the quantum mechanical phenomenon where a wavefunction can propagate through a potential barrier.

Uses of quantum tunneling:

- quantum tunneling composites (touch screen technology)

- quantum cryptography

- scanning tunnelling microscope

What is Quantum Tunneling?
Video explanation
Highly recommend downloading this app to understand quantum physics well.

12.2 Nuclear physics

Rutherford scattering and nuclear radius

The Rutherford experiment (carried out by Geiger and Marsden) proved that every atom has a nucleus where all of its positive charge and most of its mass is concentrated.

The experimenters expected that the beam of alpha particles will go through the gold leaf, however the dense positive charge inside the gold atoms repelled the alpha particles, which made them deflect and hit the sides of the screen at different angles.

The experiment allows one to derive the nuclear radius for an atom:

\begin{gather} R=R_{0} A^{\frac{1}{3}}\\ \notag\\ R=nuclear\ radius,\ R_{0} =Fermi\ radius\ \left( 1.2\times 10^{-15} m\right) \notag\\ A=nucleon\ number \notag \end{gather}
Rutherford Experiment demonstration
Video explanation

Nuclear density

Since the nucleus is spherical, the density of nuclear material can be calculated with:

\begin{gather} \rho =\frac{M}{V} =\frac{3u}{4\pi R^{\ 3}_{0}}\\ \notag\\ \rho =density,\ M=mass,\ V=volume,\ u=uniform\ atomic\ mass \notag\\ R^{\ }_{0} =Fermi\ radius\ \left( 1.2\times 10^{-15} m\right) \notag \end{gather}

The only object that has a nuclear density of this value is a neutron star.

Video explanation

Deviations from Rutherford scattering

When the Rutherford experiment is done using higher energy alpha particles, the scattering relationship does not agree with experimental results. At higher energies, the strong nuclear attractive forces overcome the electrostatic repulsion, as the alpha particles approach the nucleus closer.

The method of closest approach is thus only an approximation; more accurate results can be found using electron diffraction.

Electron diffraction

Electrons are not effected by the strong nuclear force, as they are leptons.

\begin{gather} sin\theta \approx \frac{\lambda }{D}\\ \notag\\ D=nuclear\ diameter,\ \lambda =de\ Broglie\ wavelenght \notag\\ \theta =angle\ between\ first\ diffraction\ minimum\ and\ straight\ through\ position \notag \end{gather}

Wavelenght of electron:

\begin{gather} \lambda =\frac{hc}{E}\\ \notag\\ E=electron\ energy,\ h=Planck's\ constant,\ c=speed\ of\ light \notag\\ \lambda =de\ Broglie\ wavelenght \notag \end{gather}

Same order of magnitude as the size of nucleus.

Using electrons of higher energies

When electrons of much greater energies than 420 MeV are used in scattering, the collisions are no longer elastic. The KE that bombarding electrons lose is converted into mass as several mesons emitted from the nucleus.

At even higher energies, the electrons penetrate deeper into the nucleus (deep inelastic scattering) and provide evidence of quarks.

Energy levels in the nucleus

The nucleus, like the atom, has discrete energy levels whose location and properties are governed by the rules of quantum mechanics.

Atoms that decay through gamma decay emit distinct frequencies of gamma rays which correspond to distinct energy levels.

Video explanation

Negative beta decay

Experiments show that beta particles emitted by a source have a continous energy spectrum and are not of discrete single energy as alpha particles or gamma photons.

An explanation for this is that a third particle is emmited, in order to conserve mass, energy, momentum and spin. That particle is the neutrino, or in this case the electron antineutrino.

What is a Neutrino?
Video explanation

The law of radioactive decay

Radioactive decay is a random and unpredictable process.

The probability that an individual nucleus will decay in a given time interval is known as the decay constant (λ).

The activity of the sample (A) is the number of nuclei decaying in a second. It is measured in becquerel (Bq).

\begin{gather} A=\lambda N\\ \notag\\ A=activity,\ N=number\ of\ undecayed\ nuclei \notag\\ \lambda =decay\ constant \notag \end{gather}

This can be rewritten as:

\begin{gather} A=\lambda N_{0} e^{-\lambda t}\\ \notag\\ A=activity\ of\ sample,\ \lambda =decay\ constant,N_{0} =initial\ number\ of\ nuclei, \notag\\ \ t=time \notag \end{gather}

Measuring long half-lives

Some nuclides have long half lives, like uranium-238 (4.5 billion years). For that a pure sample of the nuclide needs to be separated. From that the activity can be measured by multiplying count rate by the ratio:

\begin{equation} \frac{area\ of\ sphere\ of\ radius\ equal\ to\ the\ position\ of\ the\ G-M\ tube\ window}{are\ of\ G-M\ tube\ window} \end{equation}
Why don't all heavy elements decay to Fe56? (optional)

Topic 12 Problems

1. Light of wavelength λ and intensity I is incident on a metallic surface and electrons are emitted. Which of the following change will, by itself, result in a greater number of electrons being emitted?

A. increase λ

B. decrease λ

C. increase I

D. decrease I

2. The absorption spectra of hydrogen atoms at a low temperature and at a high temperature are compared. What is the result of such a comparison?

A. The spectra are identical

B. The spectra are identical but the high temperature spectrum is more prominent

C. There are more lines in the absorption spectrum at low temperature

D. There are more lines in the absorption spectrum at high temperature

3. Two beams of electrons of the same energy approach two potential barriers of the same height. One barrier has width w and the other width 2w. Which is a correct comparison of the tunnelling probability and the de Broglie wavelength of the transmitted electrons in the two beams?





4. The graph shows the wavefunction of a particle. Near which position is the particle most likely to be found?





5. Which is evidence for the existence of nuclear energy levels?

A. The short range of the nuclear force

B. The energies of alpha and gamma particles in radioactive decay

C. The energies of beta particles in radioactive decay

D. The emission spectra of gases at low pressure

6. The activity of a pure radioactive sample is A when the number of nuclei present in the sample is N. Which graph shows the variation of A with N?





7. Alpha particles of energy E are directed at a thin metallic foil made out of atoms of atomic number Z. For what values of E and Z are deviations from Rutherford scattering likely to be observed?





8. According to the de Broglie hypothesis, matter waves are associated with

A. electrons only

B. charged particles only

C. neutral particles only

D. all particles

9. Light incident on a clean metal surface produces photoelectrons. The threshold frequency of the light is determined by

A. the intensity of the incident light

B. the wavelength of the incident light

C. the nature of the metal surface

D. the maximum kinetic energy of the photoelectrons

10. Which one of the following best shows the variation with kinetic energy E of the de Broglie wavelength λ associated with a particle?





Number of correct answers: