black body radiation hot filament glows. classical physics cant explain the observed wavelength...
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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.