Light & Matter Interactions

Spectral Lines

William Hyde Wollaston [1766 - 1828] generated a solar spectrum using a prism, and noticed black lines in some sections. "The line A that bounds the red side of the spectrum is somewhat confused, which seems in part owing to want of power in the eye to converge red light. The line B, between red and green, in a certain position of the prism, is perfectly distinct; so also are D and E, the two limits of violet. But C, the limit of green and blue, is not so clearly marked as the rest; and there are also, on each side of this limit, other distinct dark lines, / and g, either of which, in an_ imperfect experiment, might be mistaken for the boundary of these colours."

Reproduced from Philosophical Transactions of the Royal Society of London, vol. 92 (1802), p. 380 (Plate XIV).

Joseph con Fraunhofer [1787-1826] rediscovered the lines 15 years after Wollaston. If the physical origins of the lines were understood, then we could learn more about the sun and what makes it work.

Denkschriften der K. Acad. der Wissenschaften zu München 1814-15, pp. 193-226.

Different types of spectra: continuous, emission, and adsorption.

Kirchoff's Laws

A hot, dense gas or hot solid object produces a continuous spectrum with no dark spectral lines
A hot, diffuse gas produces bright spectral lines (emission lines)
A cool, diffuse gas in front of a source of a continuous spectrum produces dark spectral lines (absorption lines) in the continous spectrum.

Inside atoms

$$\begin{equation} \frac{1}{\lambda} = R_H \left( \frac{1}{m^2} - \frac{1}{n^2} \right) \end{equation}$$

If we let $m = 2$ in the above equation, then setting $n = 3,4,\ldots,6$ yields the following wavelengths for the expected emissions.

n	$\lambda$ [nm]
3	656.3
4	486.1
5	434.0
6	410.2
7	397.0

Classical Atoms

According to Maxwell's laws of EM, a charge moving in a circle should radiate energy because of the centripetal acceleration.

The Bohr Model

The Bohr Radius

Starting with Coulomb's Law and $F = ma$: $$\begin{equation} -\frac{(Ze)e}{4 \pi \epsilon_0 r^2} = - \frac{m_e v^2}{r} \end{equation}$$ Then taking the kinetic energy of the orbiting particle: $$\begin{equation} K = \frac{1}{2}m_e v^2 = \frac{Ze^2}{8 \pi \epsilon_0 r} \label{eq:kineticenergyelectron} \end{equation}$$ The potential energy is $$\begin{equation} U = - \frac{Ze^2}{4 \pi \epsilon_0 r} \end{equation}$$ Summing the kinetic and potential to get the total energy: $$\begin{equation} E = K + U = \frac{Ze^2}{8 \pi \epsilon_0 r}-\frac{Ze^2}{4 \pi \epsilon_0 r} = -\frac{Ze^2}{8 \pi \epsilon_0 r} \end{equation}$$

The Bohr model suggests that instead of the total energy being capable of taking on any value, instead it can only take on quantized integer multiples of a constant value. Or, in terms of the angular momentum: $$\begin{equation} L = m_e v r = \frac{n h}{2 \pi} = n \hbar \end{equation}$$ Solving for $v$ and combining this with the kinetic energy from equation \eqref{eq:kineticenergyelectron} we'll obtain for the radius, $r$: $$\begin{equation} r_n = \frac{4 \pi \epsilon_0 \hbar^2}{Z e^2 m_e} n^2 = \frac{5.29 \times 10^{-11} \; \textrm{m}}{Z}n^2 \end{equation}$$ Here, $n$ is called a quantum number and can take integer values starting with 1. $Z$ is the number of protons in the nucleus. If $n = 1$ and $Z = 1$, which corresponds to the lowest energy level of the hydrogen atom, $$\begin{equation} r_\textrm{Bohr} =0.529 \times 10^{-11} \; \textrm{m} = 0.529 \unicode{x212B} \end{equation}$$ ($1 \unicode{x212B} = $ one tenth of a nanometer or $1 \times 10^{-10} \; \textrm{m}$)

Lowest energy

The energy levels of the hydrogen atom.

$E_1$ is the lowest energy of the hydrogen atom: -13.6 eV.

Now, if we are interested in the difference between two energies levels of an electron: $$\begin{equation} \Delta E = E_\rm{high} - E_\rm{low} = E_\rm{photon} \end{equation}$$

Spectrum Lines

Find the wavelength of a photon emitted during a transition from $n = 3$ to $n=2$ for an electron in a hydrogen atom.

Kirchoff's laws, again

A hot gas, or hot solid produces blackbody radiation. This is a continuous spectrum that follows Plancks Law for $B_\lambda(T)$.
A hot diffuse gas produces emission lines caused by energy lost by electrons as they go from a higher level to a lower level and emit photons at the corresponding frequencues.
A cool diffuse gas in front of a source of continuous spectrum will have dark lines when an electron gets excited by incoming energy.

Quantum Theory

The Photoelectric Effect

Shining light on a metal can knock electrons out of their atomic orbitals and enable them to become free electrons, and thus produce a current. If you think about this phenomenon based on classical physics, it would seem likely that the amount of energy in each of these electrons would be dependent on the brightness of the light. However, experiments showed that it didn't depend on the brightness of the light. Increasing the intensity of the light does increase the number of electrons produced, but not their kinetic energy. It turned out that the frequency of the light, or its color, was instead correlated with the energy of the electrons produced.

Each of the metals will have a cut-off frequency that determines how much energy is needed to knock an electron out of the orbit. Thus, the light hitting the metal will only be able to eject electrons if its frequency is above a certain amount. This is the photoelectric effect in a nutshell.

This was a mystery because the classical Maxwell equations didn't imply that different colors have different energies. Einstein solved this problem by suggesting that light traveled as photons, each with a definite energy given by: $$E_\textrm{photon} = h \nu = \frac{h c}{\lambda}$$

Modern terminology call this the work function, $\phi$ of the metal: $$ \phi = h f_0 $$ where $h$ is the Planck Constant, and $f_0$ is the threshold frequency of the material.

Photons

The energy of a single, reddish, photon: $$\begin{equation} E_\textrm{photon} = h\nu = \frac{hc}{\lambda} \simeq \frac{1240 \; \textrm{eV nm}}{700 \; \textrm{nm}} = 1.77 \; \textrm{eV} \end{equation}$$ is about half that of a blue photon: $$\begin{equation} E_\textrm{photon} = h\nu = \frac{hc}{\lambda} \simeq \frac{1240 \; \textrm{eV nm}}{400 \; \textrm{nm}} = 3.10 \; \textrm{eV} \end{equation}$$

While the spectroscopy experiments were made possible by the wave nature of light, a full explanation of what caused the dark (or bright) lines would require using the particle description of light: the photon.

*Note on the electron-volt: remember the definition of a volt: the potential difference that will give a 1 Coulomb charge 1 Joule of energy if it passes through that difference. The units are: $$\begin{equation*} V = \frac{\textrm{Potential Energy}}{\textrm{Charge}} \end{equation*}$$ So if we multiply the Volt times a Coulomb, we'll obtain a unit of energy, and that's exactly what an electron volt is: $$\textrm{an electronvolt} = \; \textrm{Volt} \times \; \textrm{Charge} = \; \textrm{energy}$$ It's the energy gained or lost by an electron as it moves across 1 volt or potential difference. $$1 \; \textrm{eV} = 1 \; \textrm{V} \times 1.6021766208(98) \times 10^{−19} \; \textrm{C}$$ It's a very small amount of energy, but useful when dealing with photons. Some of the other physical constants can be expressed in units of eV rather than Joules:

Constant	symbol	SI	eV
Planck's Constant	$h$	6.626 × 10^-34 J s	4.1357 × 10^-15 eV s
Reduced	$\hbar$	1.055 × 10^-34 J s / rad	6.582 × 10^-16 eV s / rad
	$h c$	1.99 × 10^-25 J m	1240 eV nm

de Broglie wavelength

$$\begin{equation} \lambda = \frac{h}{p} \end{equation}$$

Uncertainty Principle

$$\begin{equation} \Delta x \Delta p \ge \frac{1}{2}\hbar \end{equation}$$

Other forms of the uncertainty Principle

$$\begin{equation} \Delta E \Delta t \approx \hbar \end{equation}$$

Tunneling

Tunneling through a barrier.