The need for quantization - Quantum Mechanics for Chemistry

What is the nature of light?¶

Figure 1:Electromagnetic radiation has perpendicular electric and magnetic components that propagate at the speed of light. Unlike other waves (water, sound), light needs no medium and can travel in vacuum.

According to classical wave theory, light is seen as a traveling wave consisting of electric and magnetic components.
We will soon see that this picture of light as an electromagnetic wave is not the whole story, and radically new ideas are needed to understand a wide variety of phenomena involving the interaction of light with atoms and molecules.

Spectrum of electromagnetic waves¶

Visible light occupies a narrow frequency region in between.
High-frequency waves carry much higher energy. This means X-rays or gamma rays can only be generated by heating “stuff” up to very high temperatures. This happens naturally at the core of the Sun!
Low-frequency waves carry less energy. They can be generated in a “microwave” or by broadcasting antennas.
So what is the relationship between the frequency of radiation $\nu$ and its energy $E$ ? This is not such a trivial question. In fact, this very question arose in connection with black body radiation, an experiment that forever changed the course of history by giving birth to quantum mechanics!

Relationship between frequency, wavelength and speed of light.¶

Figure 3:Definitions of wavelength $\lambda$ and frequency $\nu$ .

Black body as a model for heated objects.¶

Figure 4:A guide to black body radiation from PhD Comics.

Watch this beautiful animation; the first 3 minutes focus solely on blackbody radiation.

Black body as an idealized model¶

This model is called a black body because it absorbs every wavelength that hits its surface, therefore appearing as a perfectly black object.
If an object has a color, it is because it reflects certain wavelengths of light, which are then detected by the retina of our eye. The distribution of wavelengths emitted by a black body is determined only by its temperature!

Figure 5:When heating up a material we observe three things: (1) the radiation intensity increases, implying higher radiated energy; (2) the distribution of emitted wavelengths shifts to lower values, or equivalently the frequency distribution shifts to higher values; and (3) the color of the material changes from red to yellow to blue.

Ultraviolet catastrophe of classical mechanics¶

What is radiation in classical mechanics? Radiation is considered a wave with frequency $\nu$ . In a heated body, naturally vibrating springs (which represent atoms or molecules) generate waves with the same frequency.

Packing wave modes in a box. One can fit more high-frequency (short-wavelength) waves in the box than low-frequency ones. The number of waves we can fit in a cubic box in the frequency region $[\nu, \nu+d\nu]$ can be estimated to be $\sim d \nu^3 \sim \nu^2 d\nu$ . The constant of proportionality requires a few more steps to derive, which we skip, writing only the final result:

dN_{\nu} = \frac{8\pi}{c^3} \cdot \nu^2 d\nu

(2)

Equipartition of energy. From thermodynamics we know that in equilibrium each degree of freedom, or each oscillator, gets the same energy $k_BT$ , where $k_B$ is the Boltzmann constant.

\langle E\rangle = k_BT

(3)

Every vibrating spring in a heated body thus has the same energy regardless of frequency. Think about this assumption for a second!
Radiation energy distribution. The distribution of radiated frequencies is then the product of the average thermal energy and the number of springs in a frequency interval:

\rho({\nu}) = k_B T \cdot \frac{8\pi}{c^3}\nu^2

(4)

Ultraviolet catastrophe. The energy distribution shoots to infinity at high $\nu$ (or low $\lambda$ ). This is known as the ultraviolet catastrophe! Integrating $\rho$ over all frequencies gives the total amount of radiation, which in this case is infinite. A light bulb could destroy the universe! Something is off with our classical prediction.

Max Planck and the trick of quantization¶

In 1900 Planck found that the theoretical curve can very closely match the experimental curve if one postulates that only discrete (quantized) values of energy are possible.

This means atoms and molecules absorb and emit radiation in discrete quantities, multiples of $h\nu$ , which are called quanta!
When light is emitted or absorbed, the atom or molecule jumps from one state to another and the energy difference $h\nu$ is either coming from light or is used to generate light.
Note how small $h$ is in the macroscopic units (such as J s). This is why quantization of energy is hardly noticeable and classical mechanics works so well at the macro scale. In the limit $h \rightarrow 0$ , $E$ becomes continuous, and an arbitrary real value of E is allowed. This is the classical limit.

The black body radiation distribution function¶

Deriving Black Body radiation formula

Planck hypothesized that the energy of oscillators in a black body is quantized and given by:

E_n = n h \nu

(6)

where $n$ is a positive integer, $h$ is Planck’s constant, and $\nu$ is the frequency.

The average energy of an oscillator is found by summing over all possible energies, weighted by the Boltzmann factor:

\langle E \rangle = \frac{\sum_n E_n e^{-E_n/kT}}{\sum_n e^{-E_n/kT}}

(7)

Substituting $E_n = n h \nu$ , the sum becomes:

\langle E \rangle = \frac{\sum_n n h \nu e^{-n h \nu / kT}}{\sum_n e^{-n h \nu / kT}}

(8)

This sum is a geometric series. For the geometric series of the form:

S = \sum_{n=0}^{\infty} x^n

(9)

The sum is given by:

S = \frac{1}{1 - x} \quad \text{for} \quad |x| < 1

(10)

In the context of Planck’s derivation, we use the series:

\sum_{n=0}^{\infty} e^{-n h \nu / kT}

(11)

This series can be summed as:

\sum_{n=0}^{\infty} e^{-n h \nu / kT} = \frac{1}{1 - e^{-h \nu / kT}}

(12)

The series involving $n$ in the numerator is:

\sum_{n=0}^{\infty} n e^{-n h \nu / kT}

(13)

This can be evaluated using the derivative with respect to $x$ :

\sum_{n=0}^{\infty} n x^n = x \frac{d}{dx} \left( \frac{1}{1 - x} \right) = \frac{x}{(1 - x)^2}

(14)

Substituting $x = e^{-h \nu / kT}$ , we get:

\sum_{n=0}^{\infty} n e^{-n h \nu / kT} = \frac{e^{-h \nu / kT}}{(1 - e^{-h \nu / kT})^2}

(15)

Using these results, Planck’s formula for the average energy becomes:

\langle E \rangle = \frac{h \nu}{e^{h \nu / kT} - 1}

(16)

The energy density $\rho(\nu, T)$ is then obtained by multiplying the average energy by the density of states and the number of oscillators per unit volume:

\rho(\nu, T) = \frac{8 \pi \nu^2}{c^3} \cdot \frac{h \nu}{e^{h \nu / kT} - 1}

(17)

This is Planck’s law, which describes the spectral density of radiation emitted by a black body in thermal equilibrium at a temperature $T$ .

Assuming that the energy of an oscillator is quantized, Planck derived a new expression for the average energy which, unlike the classical expression $k_BT$ , now depends on the frequency of oscillation:

\langle E \rangle = \Big[ \frac{h\nu}{e^{\frac{h\nu}{ kT}} - 1}\Big]

(18)

With this expression we end up with a distribution of oscillator energies that tends to zero in the high-frequency limit.

\rho_{\nu}(T) = \frac{8\pi \nu^2}{c^3} \cdot \Big[\frac{1}{e^{\frac{h\nu}{kT}} - 1} \Big]

(19)

You can also express the distribution in terms of wavelength by making the substitution $\nu = c/\lambda$ :

\rho_{\lambda}(T) = \frac{8 \pi hc}{\lambda^5} \cdot \Big[ \frac{1}{e^{\frac{hc}{\lambda kT}} - 1}\Big]

(20)

The expressions $\rho_{\lambda}(T)d\lambda$ and $\rho_{\nu}(T)d\nu$ have units of energy per volume, which is why they are often referred to as the energy density of radiation. By integrating over the entire spectrum (e.g., all frequencies or wavelengths) we obtain the total energy of radiation per volume!

\int^{\infty}_0 \rho_{\nu}(T)d\nu = \sigma T^4

(21)

$\sigma=5.6697 \cdot 10^{-8} J m^{-2} K^{-4} s^{-1}$ is called the Stefan-Boltzmann constant.

In some books you may also find black body radiation characterized via the radiation flux, measured per unit wavelength and per unit solid angle: $B_{\lambda} = \frac{c}{4\pi} \cdot \rho_{\lambda}$

Wien’s displacement law¶

Connecting the temperature of a black body with wavelength or frequency. The energy density peaks at a wavelength $\lambda_{max}$ that is inversely proportional to the temperature.
This relationship is described by Wien’s displacement law. You can derive it by setting the derivative $d\rho(\lambda)/d\lambda=0$ , which gives the wavelength at the peak of the distribution.

Explore black body radiation¶

Applications of Black Body radiation¶

planets — Figure 8:The black body is used as a standard against which the absorption of real bodies is compared. To a good approximation, stars radiate like black bodies, so we can use blackbody radiation as a model to infer the temperatures of stars from their colors. Find out more in this video on Visible Light Waves.

Rayleigh Scattering and the Color of the Sky¶

Problems¶

Conceptual question¶

Problem 1¶

Problem 2¶

Calculating the Wavelength of Radiation Using Planck’s Equation

For a monochromatic (means single wavelength) radiation with an energy of $3.5 \, \text{eV}$ calculate the wavelength. Use Planck’s equation to relate the energy of radiation to its wavelength. The values of constants are:

Planck’s constant, $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$
Speed of light, $c = 3.00 \times 10^8 \, \text{m/s}$
$1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$

Solution:

First, convert the energy of the radiation from electron volts (eV) to joules (J):

E = 3.5 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 5.607 \times 10^{-19} \, \text{J}

(23)

Now, use Planck’s equation to relate the energy $E$ of the photon to its wavelength $\lambda$ :

E = \frac{hc}{\lambda}

(24)

Rearranging to solve for $\lambda$ :

\lambda = \frac{hc}{E}

(25)

Substitute the known values:

\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{5.607 \times 10^{-19} \, \text{J}}

(26)

\lambda = \frac{1.988 \times 10^{-25} \, \text{J} \cdot \text{m}}{5.607 \times 10^{-19} \, \text{J}} \approx 3.55 \times 10^{-7} \, \text{m} = 355 \, \text{nm}

(27)

The wavelength of the radiation is approximately $355 \, \text{nm}$ , which is in the ultraviolet range of the electromagnetic spectrum.

Problem 3¶

Question: Wien’s displacement law

Using Wien’s displacement law, determine the wavelength $\lambda_{\text{max}}$ at which the spectral radiance of a blackbody is maximized. Given the temperature $T$ of the blackbody, calculate $\lambda_{\text{max}}$ for $T = 5800 \, \text{K}$ , which is approximately the temperature of the Sun’s surface.

Solution:

Wien’s displacement law states that the wavelength at which the spectral radiance of a blackbody peaks is inversely proportional to the temperature:

\lambda_{\text{max}} = \frac{b}{T}

(28)

where $b = 2.897 \times 10^{-3} \, \text{m} \cdot \text{K}$ is Wien’s displacement constant.

For $T = 5800 \, \text{K}$ :

\lambda_{\text{max}} = \frac{2.897 \times 10^{-3} \, \text{m} \cdot \text{K}}{5800 \, \text{K}} = 5.0 \times 10^{-7} \, \text{m} = 500 \, \text{nm}

(29)

So, the peak wavelength $\lambda_{\text{max}}$ for a blackbody at $5800 \, \text{K}$ is $500 \, \text{nm}$ , which is in the visible range.

Reference Table of Constants¶

Constant	Symbol	Value
Speed of light	$c$	$3.00 \cdot 10^8,\, {m/s}$
Planck’s constant	$h$	$6.626 \cdot 10^{-34},\, {J·s}$
Boltzmann constant	$k_B$	$1.381 \cdot 10^{-23},\, {J/K}$
Stefan–Boltzmann constant	$\sigma$	$5.67 \cdot 10^{-8},\, {W·m}^{-2}{K}^{-4}$
Wien’s displacement constant	$b$	$2.898 \cdot 10^{-3},\, {m·K}$