Black Body radiation

Hot filament glows. Classical physics cant

explain the observed wavelength distribution of EM radiation from such a hot object.

This problem is historically the problem that leads to the rise of quantum physics during the turn of 20th century

Need for Quantum Physics Problems remained from classical mechanics

that relativity didn’t explain Attempts to apply the laws of classical physics

to explain the behavior of matter on the atomic scale were consistently unsuccessful

Problems included: blackbody radiation

The electromagnetic radiation emitted by a heated object photoelectric effect

Emission of electrons by an illuminated metal

Quantum Mechanics Revolution Between 1900 and 1930, another revolution

took place in physics A new theory called quantum mechanics was

successful in explaining the behavior of particles of microscopic size

The first explanation using quantum mechanics was introduced by Max Planck Many other physicists were involved in other

subsequent developments

Blackbody Radiation An object at any temperature is known

to emit thermal radiation Characteristics depend on the temperature

and surface properties The thermal radiation consists of a

continuous distribution of wavelengths from all portions of the em spectrum

Blackbody Radiation, cont. At room temperature, the wavelengths of the

thermal radiation are mainly in the infrared region

As the surface temperature increases, the wavelength changes It will glow red and eventually white

The basic problem was in understanding the observed distribution in the radiation emitted by a black body Classical physics didn’t adequately describe the

observed distribution

Blackbody Radiation, final A black body is an ideal system that

absorbs all radiation incident on it The electromagnetic radiation emitted

by a black body is called blackbody radiation

Blackbody Approximation A good approximation of a

black body is a small hole leading to the inside of a hollow object

The hole acts as a perfect absorber

The nature of the radiation leaving the cavity through the hole depends only on the temperature of the cavity

Blackbody Experiment Results The total power of the emitted radiation

increases with temperature Stefan’s law (from Chapter 20):

P = AeT4

The peak of the wavelength distribution shifts to shorter wavelengths as the temperature increases Wien’s displacement law maxT = 2.898 x 10-3 m.K

Real life blackbody A close

approximation of blackbody radiator

The colour of the light emitted from the charcoal depends only upon the temperature

This figure shows two stars in the constellation Orion. Betelgeuse appears to glow red, while Rigel looks blue in color. Which star has a higher surface temperature?

(a) Betelgeuse

(b) Rigel

(c) They both have the same surface temperature.

(d) Impossible to determine.

example

Stefan’s Law – Details P = AeT4

P is the power is the Stefan-Boltzmann constant

= 5.670 x 10-8 W / m2 . K4

Stefan’s law can be written in terms of intensity I = P/A = T4

For a blackbody, where e = 1

Wien’s Displacement Law

maxT = 2.898 x 10-3 m.K max is the wavelength at which the curve

peaks T is the absolute temperature

The wavelength is inversely proportional to the absolute temperature As the temperature increases, the peak is

“displaced” to shorter wavelengths

Intensity of Blackbody Radiation, Summary

The intensity increases with increasing temperature

The amount of radiation emitted increases with increasing temperature The area under the

curve The peak wavelength

decreases with increasing temperature

Exmple Find the peak wavelength of the

blackbody radiation emited by (A) the Sun (2000 K) (B) the tungsten of a lightbulb at 3000 K

Solutions (A) the sun (2000 K) By Wein’s displacement law,

(infrared) (B) the tungsten of a lightbulb

at 3000 K

Yellow-green

3

max2.898 10 m K

2000K1.4 m

3

max2.898 10 m K

5800K0.5 m

Rayleigh-Jeans Law An early classical attempt to explain

blackbody radiation was the Rayleigh-Jeans law

At long wavelengths, the law matched experimental results fairly well

4

2I , Bπck T

λ Tλ

Rayleigh-Jeans Law, cont. At short wavelengths,

there was a major disagreement between the Rayleigh-Jeans law and experiment

This mismatch became known as the ultraviolet catastrophe

You would have infinite energy as the wavelength approaches zero

Max Planck Introduced the

concept of “quantum of action”

In 1918 he was awarded the Nobel Prize for the discovery of the quantized nature of energy

Planck’s Theory of Blackbody Radiation In 1900 Planck developed a theory of

blackbody radiation that leads to an equation for the intensity of the radiation

This equation is in complete agreement with experimental observations

He assumed the cavity radiation came from atomic oscillations in the cavity walls

Planck made two assumptions about the nature of the oscillators in the cavity walls

Planck’s Assumption, 1 The energy of an oscillator can have only

certain discrete values En

En = nhƒ n is a positive integer called the quantum number h is Planck’s constant ƒ is the frequency of oscillation

This says the energy is quantized Each discrete energy value corresponds to

a different quantum state

Planck’s Assumption, 2 The oscillators emit or absorb energy

when making a transition from one quantum state to another The entire energy difference between the

initial and final states in the transition is emitted or absorbed as a single quantum of radiation

An oscillator emits or absorbs energy only when it changes quantum states

Energy-Level Diagram An energy-level diagram

shows the quantized energy levels and allowed transitions

Energy is on the vertical axis

Horizontal lines represent the allowed energy levels

The double-headed arrows indicate allowed transitions

More About Planck’s Model The average energy of a wave is the average

energy difference between levels of the oscillator, weighted according to the probability of the wave being emitted

This weighting is described by the Boltzmann distribution law and gives the probability of a state being occupied as being proportional to

where E is the energy of the state BE k Te

Planck’s Model, Graphs

Planck’s Wavelength Distribution Function Planck generated a theoretical

expression for the wavelength distribution

h = 6.626 x 10-34 J.s h is a fundamental constant of nature

2

5

2

1I ,

Bhc λk T

πhcλ T

λ e

Planck’s Wavelength Distribution Function, cont. At long wavelengths, Planck’s equation

reduces to the Rayleigh-Jeans expression

At short wavelengths, it predicts an exponential decrease in intensity with decreasing wavelength This is in agreement with experimental

results

Example: quantised oscillator vs classical oscillator

A 2.0 kg block is attached to a massless spring that has a force constant k=25 N/m. The spring is stretched 0.40 m from its EB position and released.

(A) Find the total energy of the system and the frequency of oscillation according to classical mechanics.

Solution (A)

Mechanical Energy, E = ½ kA2 = …= 2.0 J Frequency, 1

... 0.56Hz2

kf

m

(B) Assuming that the energy is quantised, find the quantum number n for the system oscillating with this amplitude.

QUICK QUIZ 40.1 (For the end of section 40.1) In a certain experiment, you pass a

current through a wire and measure the spectrum of light that is emitted from the glowing filament. For a current I1, you measure the wavelength of the highest intensity (also called max) to be 1. You then increase the voltage across the wire by a factor of 8 and the current increases by a factor of 2 to 2I1. The wavelength of the highest intensity will shift to a) 41, b) 21, c) 21, d) 1/ 2, e) 1/ 2, or f) 1/ 4.

QUICK QUIZ 40.1 ANSWER

(e). Assuming that the wire behaves like a blackbody, the wavelength with the highest intensity will be inversely proportional to the absolute temperature according to Wien’s displacement law, or max 1/T. In addition, the power radiated from the wire is proportional to the absolute temperature to the fourth power; from Stefan’s law, or P T4. Also, the power dissipated in the wire is given by P = IV. Combining these equations, one finds that,

.2

1

)82(

1

)(

1114/14/14/1max

IVPT

QUICK QUIZ 40.2

The oscillators in a blackbody may oscillate a) at only certain frequencies, b) with only certain energies, c) at only certain frequencies and with only certain energies, d) with only certain energies for a certain frequency, or e) at only certain frequencies for a certain energy.

QUICK QUIZ 40.2 ANSWER

(d). The condition that the oscillators in a blackbody can have only discrete energy values according to Equation 40.4, En = nhf, does not imply that the oscillator may only oscillate at discrete frequencies. In fact, the blackbody spectrum is continuous precisely because all oscillation frequencies are allowed. What Equation 40.4 does imply is that if an oscillator is oscillating at a specific frequency, then it may only occupy certain discrete energy states, and that transitions are only allowed between these discrete energy states.