Experimental basis of QM

A set of results that spawned 20th century physics

Experimental basis of QM

Overview

Discovery of the X-ray and the electron

Determination of the quantised charge of the electron

Emission Spectra

Blackbody radiation

The photoelectric effect

Compton effect

Electron charge to mass ratio

Experimental basis of QM

Setting the scene

Experimental basis of QM

Background

Electromagnetic waves

Experimental basis of QM

Status quo

Recall Maxwell's equations

Experimental basis of QM

Status quo

Electromagnetic waves

we can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena

c =\frac{1}{\sqrt{{\epsilon_0}{\mu_0}}}

Lecture at Kings College (1862)

This velocity is so nearly that of light, that it seems we have strong reason to conclude that light itself (including radiant heat, and other radiations if any) is an electromagnetic disturbance in the form of waves propagated through the electromagnetic field according to electromagnetic laws.

A Dynamical Theory of the Electromagnetic Field (1864) Introduction, p. 466, §20.

Experimental basis of QM

Status quo

Electromagnetic waves

c =f\lambda

Experimental basis of QM

Background

Emission spectra

Experimental basis of QM

Line Spectra

Characteristic line spectra of elements

Experimental basis of QM

Line Spectra

Balmer series (1885)

\lambda = 364.6 \text{nm} \left(\frac{k^2}{k^2-4}\right)

Empirically, Balmer found an expression for the pattern of wavelengths in the Hydrogen spectrum

\text{where} \quad k=3,4,5, ...

\text{e.g.}\quad k=3 \implies\lambda = 364.6 \text{nm} \left(\frac{3^2}{3^2-4}\right)=656.3\text{nm}

Experimental basis of QM

Line Spectra

Rydberg and Ritz (1890)

\frac{1}{\lambda} = \frac{1}{364.6 \text{nm}} \left(\frac{k^2-4}{k^2}\right)=\frac{4}{364.6 \text{nm}} \left(\frac{1}{2^2}-\frac{1}{k^2}\right)

Rydberg and Ritz recognized that Balmer's result can be expressed as:

\frac{1}{\lambda} = R_H\left(\frac{1}{n^2}-\frac{1}{k^2}\right)

Which generalizes to

\text{where} \quad R_H=1.0968 \times10^7 m^{-1}

Experimental basis of QM

Background

Cathode Rays

Experimental basis of QM

Discovery of the X Ray and the Electron

Rontgen's (accidental) discovery of X-rays (1895)

From T & R, p 86

Experimental basis of QM

Discovery of the X Ray and the Electron

Thomson's investigation of Cathode rays

Experimental basis of QM

Background

charge to mass ratio

Experimental basis of QM

Discovery of the X Ray and the Electron

Thomson's experiment -- electron charge to mass ratio (1897)

From T & R, p 86

J.J. Thomson showed that cathode-rays were charged particles by showing their deflection in magnetic/electric fields.

Furthermore, he managed to measure the charge to mass ratio of the electron.

Experimental basis of QM

Discovery of the X Ray and the Electron

Thomson's experiment -- electron charge to mass ratio (1897)

From T & R, p 86

Experimental basis of QM

Background

Millikan's quantization of charge

Experimental basis of QM

Determination of the electron charge

Millikan's experiment -- electron charge (1911)

From T & R, p 89

Millikan suspended oil drops between two plates by changing the potential difference across the plates.

The equilibrium between the force of gravity and the electric force gave an estimate of the charge on the oil drop, in terms of its volume.

The volume was estimated from the terminal velocity.

Millikan found that the drops carried electric charge that was quantized!

The elementary charge, he found, was

e=1.602\times10^{-19}C

Experimental basis of QM

Determination of the electron charge

Millikan's experiment -- electron charge (1911)

Experimental basis of QM

Blackbody Radiation and the birth of the quanta

Experimental basis of QM

Blackbody Radiation

What is a blackbody?

All bodies simultaneously emit and absorb radiation.

A blackbody absorbs all radiation that is incident on it, and reflects none.

A blackbody emits thermal radiation whose properties do not depend on the material, but rather on the temperature of the body.

Experimental basis of QM

Blackbody Radiation

Spectral distribution of blackbody radiation

Experimental basis of QM

Blackbody Radiation

Spectral distribution of blackbody radiation

The wavelength @ peak intensity shifts to smaller wavelengths for bodies with higher temperature.

The intensity is the total power radiated per unit area per unit wavelength at a given temperature.

The total power (per unit area) increases dramatically with temperature.

Spectral distribution of radiation emitted from a blackbody for different blackbody temperatures.

Experimental basis of QM

Blackbody Radiation

Spectral distribution of blackbody radiation

The wavelength @ peak intensity shifts to smaller wavelengths for bodies with higher temperature.

The total power (per unit area) increases dramatically with temperature.

Wien's displacement law

\lambda_\text{max} =\frac{ 2.898\times10^{-3} \text{m}\cdot \text{K}}{T}

Q: Given this solar radiation spectrum, estimate the temperature of the surface of the sun.

Q: Estimate the peak wavelength at which a human body emits thermal radiation.

Q: What color signifies an object at a higher temperature? R or B?

Experimental basis of QM

Blackbody Radiation

Spectral distribution of blackbody radiation

The total power (per unit area) increases dramatically with temperature.

Stefan-Boltzmann law

R(T)=\sigma\ T^4

for blackbody radiation

R(T)=\int_0^\infty \mathcal{I}(\lambda,T) d\lambda

where

\sigma = 5.6705\times 10^{-8} \ \text{W}/(\text{m}^2\cdot\text{K}^4)

and

For any other body (other than a blackbody)

R(T)=\epsilon \sigma\ T^4

\epsilon

where

is the emmissivity, which depends on the surface material.

Experimental basis of QM

Spectral distribution of blackbody radiation

Empirical results

Q: Estimate the total power emitted by the sun, and the intensity of solar radiation on Earth.

Given

\text{Diameter}_\text{Sun}=1.39\times 10^9 m

\text{D}_\text{Earth - Sun}=1.48\times 10^{11} m

Experimental basis of QM

Spectral distribution of blackbody radiation

Empirical results

Q: Estimate the total power emitted by the sun, and the intensity of solar radiation on Earth.

Given

\text{Diameter}_\text{Sun}=1.39\times 10^9 m

\text{D}_\text{Earth - Sun}=1.48\times 10^{11} m

Experimental basis of QM

Spectral distribution of blackbody radiation

Empirical results

Q: Estimate the time it would take for the human body to radiate enough heat to drop from 37 degrees Celsius to 36 (in the absence of any other heat input or output.)

You will have to make some assumptions about:

effective emissivity of body surface
surface area of emitting body
mass of body
effective heat capacity of body

Approach:

Estimate how much heat needs to be radiated, and
estimate the rate at which humans radiate energy, then
combine the two estimates for an estimate of the time.

Experimental basis of QM

Quantization of thermal energy spectrum

Spectral distribution of blackbody radiation

The culmination of classical physics attempting to describe thermal radiation of a blackbody

1905

Rayleigh-Jeans formula

\mathcal{I}(\lambda,T)=\frac{2 \pi c \ k_B T}{\lambda^4}

The ultraviolet catastrophe!

Experimental basis of QM

Quantization of Energy

Max Planck's Energy Distribution Law of thermal radiation

Experimental basis of QM

Quantization of Energy

Max Planck's quantization of energy

Blackbody radiation results from the transitions of quantized atomic oscillators.

The energy of an oscillator can only have discrete values

E_n=nhf

The transitions between the discrete energies of an oscillator are

E_\gamma=\Delta E=hf

h=6.26\times10^{-34} \text{J}\cdot\text{s}

Experimental basis of QM

Quantization of energy

Max Planck's quantization of energy

\mathcal{I}(\lambda,T)=\frac{2\pi h c^2}{\lambda^5\left( e^{hc/\lambda k_B T}-1\right)}

Experimental basis of QM

Quantization of energy

Max Planck's quantization of energy

\mathcal{I}(\lambda,T)=\frac{2\pi h c^2}{\lambda^5\left( e^{hc/\lambda k_B T}-1\right)}

Wien's displacement law

\lambda_\text{max} =\frac{ 2.898\times10^{-3} \text{m}\cdot \text{K}}{T}

Experimental basis of QM

Quantization of energy

Max Planck's quantization of energy

\mathcal{I}(\lambda,T)=\frac{2\pi h c^2}{\lambda^5\left( e^{hc/\lambda k_B T}-1\right)}

Stefan-Boltzmann law

R(T)=\sigma\ T^4

Experimental basis of QM

Quantization of energy

Max Planck's quantization of energy

\mathcal{I}(\lambda,T)=\frac{2\pi h c^2}{\lambda^5\left( e^{hc/\lambda k_B T}-1\right)}

Q: For a black body at temperature of T_0, what percentage of its emitted thermal radiation has a frequency higher than f_0?

\% \mathcal{R}(T_0)_{\lambda\gt\lambda_0}=\frac{\int_0^{\lambda_0}\mathcal{I}(\lambda,T_0)}{\int_{0}^{\infty}\mathcal{I}(\lambda,T_0)}

\lambda_0=\frac{c}{f_0}

where

Experimental basis of QM

Quantization of energy

Max Planck's quantization of energy

Experimental basis of QM

The photoelectric effect and its implications

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Classically, electrons absorb energy continuously from the incident electromagnetic waves, so you would expect:

As the intensity of the light increases, energy would be transfered at a higher rate, and therefore ejected electrons will have higher kinetic energy.
Electrons should be ejected from the metal at any incident frequency of light, as long as the intensity is high enough.
The kinetic energy of the electrons only depend on intensity, and not on the frequency of the light.

Experimental basis of QM

Quantization of light energy

The photoelectric effect

According to the quantum description, by Einstein, an incoming photon gives all its energy to a single electron, thus energy is absorbed by the metal in bundles, not continuously.

The kinetic energy of the ejected electron is dependent on the incoming photon's energy (which is proportional to its frequency), not on the intensity.
The intensity of the light determines the number of photons, and therefore the number of ejected electrons.

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Observations:

e\ \Delta V_s = K_{max}

K_{max} = hf - \phi

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Suppose a Sodium surface is illuminated with light of wavelength = 300 nm. What is the maximum kinetic energy of the ejected electrons?

Example

Experimental basis of QM

Quantization of light energy

The photoelectric effect

The threshold wavelength of potassium is 558 nm. What is the work function for potassium? What is the stopping potential when light of wavelength 400 nm is used?

Example

Experimental basis of QM

Quantization of light energy

The photoelectric effect

Experimental basis of QM

Compton's Scattering

Experimental basis of QM

Quantization of light Energy

Photon momentum

E=hf=\frac{hc}{\lambda}

Recall

E^2=p^2c^2+m^2c^4

and for photons

E=pc

But, according to Planck,

Therefore, for photons

p=h/\lambda

Experimental basis of QM

Quantization of light energy

Compton's Scattering

P_e^2=(p-p'\cos\theta)^2+(p'\sin\theta)^2

In an elastic collision between a photon and a free electron:

The conservation of momentum leads to:

x: \quad P_e \cos\phi = p-p' \cos\theta

y: \quad P_e \sin\phi = p' \sin\theta

P_e^2=p^2+p'^2-2pp'\cos\theta

Experimental basis of QM

Quantization of light energy

Compton's Scattering

P_e^2=(p-p'+m_ec)^2-m_e^2c^2

In an elastic collision between a photon and a free electron:

The conservation of momentum leads to:

E+E_0=E'+E_e

pc+m_ec^2=p'c+\sqrt{m_e^2c^4+P_e^2c^2}

P_e^2=p^2+p'^2-2pp'\cos\theta

The conservation of Energy leads to:

P_e^2=p^2+p'^2-2pp'

+2(p-p')m_ec

Experimental basis of QM

Quantization of light energy

Compton's Scattering

In an elastic collision between a photon and a free electron:

(p-p')m_ec=pp'(1-\cos\theta)

combining

\frac{m_ec}{p'}-\frac{m_ec}{p}=(1-\cos\theta)

\frac{m_ec\lambda'}{h}-\frac{m_ec\lambda}{h}=(1-\cos\theta)

The conservation of momentum leads to:

P_e^2=p^2+p'^2-2pp'\cos\theta

The conservation of Energy leads to:

P_e^2=p^2+p'^2-2pp'

+2(p-p')m_ec

Experimental basis of QM

Quantization of light energy

Compton's Scattering

\Delta \lambda = \lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta)

In an elastic collision between a photon and a free electron:

The conservation of energy and momentum lead to:

i.e. If the analysis is correct, we should find a specific correlation between the angle of ricochet and the scattered photon's wavelength.

Experimental basis of QM

Quantization of light energy

The Compton effect

Compton verified this result experimentally, finding scattered photons with specific wavelengths at specific angles.

Experimental basis of QM

Quantization of light energy

The Compton effect

Suppose an x-ray photon with wavelength of 0.0500 nm loses 4% of its energy in a collision with an electron that is initially at rest. At what angle does the photon deflect (from its original trajectory)?

Example

Experimental basis of QM

Quantization of light energy

The Compton effect

Experimental basis of QM

Conclusions

Discovery of the X-ray and the electron

Determination of the quantised charge of the electron (Millikan)

Emission Spectra

Blackbody radiation

The photoelectric effect

Compton effect

Electron charge to mass ratio (Thomson)

Each type of atom emits light with a specific set of discrete energy.

Electrons are discrete entities that carry an indivisible elementary charge

The energy emitted and absorbed by atomic resonators are quantised, where the energy of the individual quanta is proportional to the frequency of the corresponding EM wave.

Light scatters from electrons as-if it was a collision between particles.

Further evidence that light energy is quantised. Photons can be treated as bundles of energy (i.e. particle-like) that can be "absorbed" and converted to KE of electrons.