Quantum Physics

Chapter 27!

Quantum – What does that mean? The study of light at the beginning of the last century led to unanswered

questions about the nature of light. Physicists use models to describe phenomena they cannot see or measure directly.

The wave model of light was successful in predicting most observed behavior, but certain observations could not be explained using a wave model.

A new hypothesis was formed : Light can also be modeled as a particle. light is quantized: it is transferred only in discrete quantities of energy.

A quantum is a discrete unit of electromagnetic energy.

Wave – Particle Duality

Particles exhibit wave properties

Waves behave as particles

Both models may be used to understand the physical world.

Quantum Mechanics

Quantum Mechanics is the branch of physics that studies extremely small particles and their interactions with electromagnetic energy.

Calculations involving such small particles deal with probabilities rather than precise values.

Blackbody Radiation

A blackbody is a perfect absorber of electromagnetic radiation. In theory it absorbs ALL incident radiation.

A box with a very small hole acts like a blackbody because nearly all incident radiation is absorbed.

As a perfect absorber, it is also a perfect emitter of radiation. We study the emissions – intensity vs. temperature

Blackbody Radiation At normal temperatures, radiation

is emitted in the infared region.

As temperature increases, the object emits radiation in the visible portion of the spectrum.

Though there is a continuous spectrum of emission wavelengths, the maximum intensity occurs at a particular wavelength.

This wavelength decreases as temperature increases.

Wein’s Displacement Law

As temperature increases, more radiation is emitted at every wavelength.

Also the wavelength of max intensity, λmax, becomes shorter.

Wein’s Law:

λmaxT = 2.90 X 10 -3 mK

Example

The visible surface of our sun is the gaseous photosphere from which radiation escapes. At the top of the photosphere the temperature is 4500 K. At a depth of 260 km the temperature is 6800 K. What are the wavelengths of maximum intensity for these temperatures? What colors do they correspond to?

Planck’s Constant – The Ultraviolet Catastrophe!

Classical physics (Newtonian) predicts that emitted energy depends on the frequency of the oscillating charge:

Intensity is proportional to (1/λ)4

i.e. As wavelength decreases, I increases.

Classical theory agrees for longer wavelengths, but for shorter wavelengths (shorter than violet) experimental data disagrees with classical physics!

Predicted intensity would grow to infinity, but experimentally the intensity drops to zero!

Planck’s Constant – The Ultraviolet Catastrophe!

Max Planck developed a theory that matched the experimental evidence, but his theory relied on the premise that energy emitted by an object can only be emitted in discrete amounts!

En = n(hf) integer multiples of hf where

h = planck’s constant = 6.63 X 10 –34 Js

The smallest amount of energy emitted by an oscillator is

E = hf. Energy is quantized!

Photoelectric Effect

One of Einstein’s famous experiments!

Light is absorbed and emitted in discrete quantities called photons with energy E = hf.

Einstein used the photon concept to explain the photoelectric effect…

Photoelectric Effect A photoelectric material is the

cathode end of a capacitor – when light shines on it, electrons can be liberated.

The capacitor is placed inside an evacuated tube.

A variable voltage supply maintains the potential across the plates.

When electrons are liberated, a current is measured in the ammeter.

Photoelectrons The energy from the light

source hitting the cathode causes electrons to be freed…

With a potential difference between anode and cathode, all ‘photoelectrons’ make it to the anode.

Current of photoelectrons can be measured, Ip

Increasing the voltage

does not increase Ip

Increasing the intensity of

light DOES increase Ip.

Photoelectrons Kinetic energy of the

photoelectrons can be measured by making V<0.

The retarding voltage repels the photoelectrons.

Photocurrent decreases as voltage gets more negative; only photoelectrons with KE > e|V| will make it to the anode.

The ‘stopping voltage’ gives the max kinetic energy of the photoelectrons

Kinetic Energy Observations

Varying the intensity changes the photoelectron current but does NOT change the stopping voltage / max KE.

Varying the frequency DOES affect the maximum kinetic energy.

Max Kinetic Energy varies linearly with increasing frequency.

There is a minimum frequency below which no current is detected.

Photoelectric Effect Explained by Einstein

hf = KE + Φ

f gives frequency of incident light

h is planck’s constant

KE is measured max kinetic energy of photoelectrons

Φ is work required to liberate an electron

KE = hf - Φ

eV or electron-volts

Small energy quantities are measured in electron-volts. An electron volt is the potential energy of an electron when pushed through a potential of 1 Volt (1 Joule/Coulomb)

1 eV = (1.6X 10 -19 C)(1 J/C) = 1.6 X 10 -19 J

Examples

The work function of a particular metal is 2.00eV. If the metal is illuminated with light of wavelength 550 nm,

a) what is the maximum kinetic energy of the emitted photoelectrons?

b) what is their maximum speed?

c) what is the stopping potential?

Cutoff Frequency or Threshold Frequency Below a particular frequency, the

photoelectrons in the photoelectric material don’t have enough energy to dislodge.

This frequency is called the cutoff frequency.

hf = KE + Φ hf0 = 0 + Φ

f0 = Φ/h gives the cutoff frequency

Example

What are the cutoff frequency and corresponding wavelength for the metal described in the last example??

Compton Scattering

Arthur Compton in 1923 described the scattering of x-rays from a carbon block by assuming incident radiation is a particle or quantum.

Compton Scattering

Observations: When monochromatic light scatters from various materials, the

wavelength of the scattered radiation is longer.

The wavelength of the scattered radiation depended upon the angle through which it was scattered.

The wavelength of scattered radiation did not depend on the nature of the scattering material.

Classical physics / wave theory predicts that the scattered radiation should have the same wavelength as the incident radiation.

Compton Scattering

Compton assumed light behaved as a particle.

He also assumed that light incident on scattering material was scattered through elastic collisions with electrons. (Why not the nucleus?)

For an elastic collision, momentum and kinetic energy are conserved.

These assumptions led Compton to a complete and accurate description of the observed scattering.

Compton’s ResultΔλ = λ’ – λ = λc(1 – cosθ)

λc = 2.43 X 10 -12 m

= Compton wavelength

Based on elastic collisions of particles, Compton derived a formula which correctly predicts the shift in wavelength.

Example

A monochromatic beam of x- ray radiation of wavelength 1.35 X 10 -10 m is scattered by the electrons in a metal foil. By what percentage is the wavelength shifted if the scattered x-rays are observed at 90 degrees?

Homework

Read carefully Section 27.4 – The Bohr Theory of the Hydrogen Atom

This week: 27.4, 28.1, 29.2 – 29.4

Do # 29, 32, 36, 38, 39, 41, 45 – 47, 50

Bohr Theory of a Hydrogen Atom

The electrons that orbit an atom are ‘allowed’ to be in distinct energy levels.

Electrons can absorb a photon of energy and jump to a higher level, or can emit a photon of energy and fall to a lower level.

Emission and Absorption Spectra

Bohr’s Model

Bohr assumed electron orbits the proton in a similar fashion to planet orbiting the sun.

He assumed the electrical force provided the centripetal force.

He also assumed that angular momentum was quantized! Could only exist in multiples of h/2π

Energy of Hydrogen Atom

Centripetal force = Electrical Force

m(v2/r) = ke2/r2 (1)

Energy = KE + PE

KE = ½ mv2; PE = -ke2/r

Energy = ke2/2r - ke2/r = ke2/2r - 2ke2/2r

Energy = -ke2/2r (2)

Energy of the Hydrogen Atom So Far: Energy = -ke2/2r (2)

So Far: This comes from classical considerations of energy and circular motion.

The Leap: Neils Bohr assumed that the angular momentum of the electron was quantized: mvr = n(h/2π) for n = 1, 2, 3…

So vn = nh/2πmrputting this into (1) and solving for r we get

rn = (h2/4π2ke2m)n2 n = 1, 2, 3, …

Energy of the Hydrogen Atom rn = (h2/4π2ke2m)n2 n = 1, 2, 3, …

gives the radius of electron orbitals

Substituting into equation (2)

En = -ke2/2r = -(2π2k2e4m/h2)(1/n2)

Hydrogen: rn = 0.0529n2 nm

En = -13.6/n2 eV

Energy of the Hydrogen Atom

n = 1 is the ground state.

n gives the principal quantum number.

ΔE = Eni – Enf

= 13.6(1/nf – 1/ni) eV

λ (nm) = hc/ ΔE

Example

What wavelength (and color) is emitted when an electron goes from n = 3 to n = 2 energy level?

Example

We know four wavelengths of the Balmer Series are in the visible range (corresponding to n = 2, 3, 4, 5, 6). There are more than just these four in the series. What type of light are they likely to be? Infared, visible or ultraviolet? What is the longest wavelength in the Balmer Series??

Homework

Do # 53, 55, 57, 59, 60, 62 – 67, 71 on page 875