Physics 361 Principles of Modern Physics Lecture 3.

Physics 361Principles of Modern Physics

Lecture 3

Blackbody RadiationAnd Its Implications on Quantum Properties

of Light

How Does Blackbody Radiation Form?Go back to an ideal gas.

All molecules move randomly.

Each one has on average a kinetic energy of .

for each mode of energy storage. “Degree of freedom.”

The higher the T, the greater the movement of molecules.

How Does Blackbody Radiation Form?Go back to an ideal gas.

Since all matter is made of charged particles, this random movement of charge should emit electromagnetic waves.

The higher the temperature, the more energy is radiated from the matter.

Remember our earlier arguments about E&M wave emitters.

EM Waves by an Antenna

• Two rods are connected to an ac source, charges oscillate between the rods (a)

• As oscillations continue, the rods become less charged, the field near the charges decreases and the field produced at t = 0 moves away from the rod (b)

• The charges and field reverse (c)• The oscillations continue (d)

So everything at a finite temperature emits light.

But why do we call them “black bodies”??

Consider Two Identical Materials at the Same Temperature

They are in thermal equilibrium, so the net energy across dashed line must be zero!

Now add a perfectly reflecting coating on only one of them.

In equilibrium the net energy across dashed line must still be zero! So the perfectly reflecting material does not emit radiation!!

Thus, perfectly reflecting bodies do not absorb or emit light.

Reflecting bodies have zero “emissitivity” or “absorptivity” – the same thing.

Something that emits a lot of light is not reflective – thus it is black!!

Blackbody Radiation – Opening in a cavity is a good approximation

Emitted light from opening should be proportional to average energy density within the enclosed cavity.

Energy density per small increment of wavelength is .

In classical wave picture of light, the number of wave excitations (modes of oscillation) grows as an inverse

power of wavelength.

That is – the smaller the wavelength, the more types of oscillations you can fit in an enclosure.

Like fitting marbles in a jar. The smaller the marbles, the more you can fit into a jar.

For wave modes, this is the same. The smaller their wavelength, the more of their mode you can fit into an enclosed space.

Mode density per small wavelength increment works out to be

*Note: we will work out details of mode counting later in this course.

So we expect more small wavelength oscillations inside the cavity. How much energy is contained in each

mode?

Classical Equipartition Theorem Stick an oscillator in container (assume motion only allowed in 1D). How much energy is on average stored in the oscillator?

must be stored in kinetic energy due to collisions.

Since oscillators vary periodically in having energy either all in kinetic or all in potential, we expect another to be stored on average in potential.

Thus, a total of kT should be stored in an oscillator mode in thermal equilibrium.

Total energy density per unit wavelength is then

This diverges as wavelength goes to zero!!

If classical blackbody radiation (“Rayleigh-Jeans” behavior) were correct... you’d need a really good

pair of sunglasses. • Everything would emit an infinite

amount of radiation (and energy).

• Raises important questions...

• Where would objects store this infinite energy source?

• Wouldn’t the world be a very bright place??

The Ultraviolet Catastrophe

• Classical theory does not match the experimental data

• At long wavelengths, the match is good

• At short wavelengths, classical theory predicts infinite energy density

• At short wavelengths, experiment shows approximately zero energy density

• This contradiction is called the ultraviolet catastrophe

Experimental Blackbody Radiation Graph

• Experimental data for distribution of energy in blackbody radiation at right.

• As the temperature increases, the total amount of energy and emitted intensity increases

– Shown by the area under the curve As the temperature increases, the

peak of the distribution shifts to shorter wavelengths (“Wien’s displacement law”)

λmax T = 0.2898 x 10-2 m • K• Total emitted intensity is proportional to T4.

Planck’s Resolution• Planck hypothesized that the blackbody radiation was produced by resonators

– Resonators were submicroscopic charged oscillators

• The resonators could only have discrete energies– En = n h ƒ

• n is called the quantum number• ƒ is the frequency of vibration• h is Planck’s constant, 6.626 x 10-34 J s

• Key point is quantized energy states• Mathematically correct, but he did not explain why this quantization occurs or

specifically where it occurs.• Gives a correct description of blackbody radiation.

Modern View – The Electromagnetic Field is Quantized into Photons

• The E&M waves have only allowed discrete allowed numbers– En = n h ƒ

• n is called the quantum number• ƒ is the frequency of vibration• h is Planck’s constant, 6.626 x 10-34 J s

• These quantum numbers are the actual number of E&M radiation particles (called “photons”) present in a mode.

• Since small wavelength excitations have very large energies, they would be unlikely to be thermally excited and there would not be kT of energy for each of these high-energy modes. This is the source of the exponential term in Plank’s description. (This can be derived using quantum statistics later in the course.)

Conclusions about E&M radiation

• Not only is the energy dependent on the frequency rather than the amplitude, but the radiation acts like discrete particles that can be counted.

• Next lecture we will explore collisions of these photons with matter to see whether they actually behave like particles.