Contents
The nucleus
Radioactivity
Radiation Measurements
Black body radiation
Statistical Mechanics
Radiation and scattering
Related topics
Related links
© The scientific sentence. 2010

Quantum theory
1. Planck postulate
Max Planck, in 1900, had known the established results related to the blackbody,
and tried then to get a conclusion. He first tried to solve the problem by a kind
of a guessing game. What could work or not to find an acceptable expression for
the spectral density energy function u(ν,T) which the experimental results (isotherms
of the blackbody) have been already established. Finally, he decided to make an ad hoc
supposition that he had made a postulate:
The energy of an oscillator can exist only in discret packets form.
The energy of a packet contains one or more quantum (plural quanta).
The number of quanta in the packet is discrete. A packet is then a quanta.
The quantum is the elementary energy quantity for an oscillator. It is equal to
hν (h is a constant and ν is the vibration frequency of the oscillator).
Then the energy an oscillator of frequency ν can be hν, 2hν, or nhν.
The energy of an oscillator is: E = E_{n} = n hν n is an integer.
That was the first statement of the quantization of energy.
It follows that the energy of an oscillator can vary (give and receive, absorb and emit)
only in discrete finite quantity form (by lumps).
Planck stated then as a principle that the energy exchange between matter (wall of
the cavity) and rays (thermal inside the cavity) is set by quanta transfert.
2. Planck law
A balckbody, at an equilibrium temperature T, emits a thermal ray inside
the vacuum of the cavity. This ray is a set of independent monochromatic
radiations with all possible frequencies. Each radiation is an electromagnetic
wave.
Let's consider one frequency ν and consider all the radiations with
this frequency. They are many. Each elecromagnetic wave (or radiation) is
assumed analogous to an oscillator with the same frequncy. According to the
Planck postulate, each oscillator has an energy in a packet form: h&n; or
2 hν ..., or nhν. Inside the set of the oscillators with the same
frequency ν, each oscillator is in a particular state corresponding
to the number of quanta that it has. If an oscillator has 0 (zero) quanta in its
packet,it is in a ground state; otherwise, it is in an excited state.
Each state is defined by the number "n" of quanta in the packet. This number "n"
characterizes the energy level of the oscillator. What's then the average enegy per
oscillator? We will use Statistics to bring an answer.
If N is the total number of the oscillators with the same frequency ν, N_{0}
the number of oscillators (with the same frequency ν) with 0(zero) quantum,
or 0 energy, ..., and N_{n} the number of oscillators (always with the
same frequency ν) with n quanta, or nhν energy, we have:
N = N_{0} + N_{1} + ... + N_{n} = ∑ N_{i} [ i: 0 → n ]
The number "n" is large, but not infinite. We will write "∞" to mention that
this number "n" is very large.
The population N_{i} of the different levels "i" of energy ihν is set
by the Boltzmann formula:
P_{i} = Probability to have N_{i} oscillators in the state "i",
then with energy E_{i} = ihν
N_{i} = N x P_{i} = exp{  E_{i}/kT}/Z
where Z is the partition function for the oscillators =
∑ exp{  E_{i}/kT} [i: 0 → ∞ ]
Similarly, we have:
P_{0} = Probability to have N_{0} oscillators in the state "0" or ground state
and N_{0} = N x P_{0}
We can then write the ratio:
N_{i}/N_{0} = exp{  (E_{i}  E_{0})/kT}= exp{  (ihν  0)/kT}
and we get the:
Boltzmann formula:
N_{n}/N_{0} = exp{  nhν/kT}
The total number N is calculated as:
N = ∑ N_{i} [ i: 0 → "∞" ]
= N_{0}∑ exp{  nhν/kT} = N_{0}∑ exp{  nhν/kT}
x =  hν/kT
∑ exp{nx} = 1 + exp{x} + exp{2x} + ... + exp{nx} + ....
= 1 + (exp{x})^{1} + (exp{x})^{2} + ... + (exp{x})^{n} + ....
= (1  (exp{x})^{n+1}) / (1  exp{x}) + ...
= (1  (exp{(n+1)x})^{n+1}) / (1  exp{x}) + ...
= (1  (exp{ (n+1)hν/kT})^{n+1}) / (1  exp{ hν/kT}) + ...
Then:
Limit ∑ exp{  nhν/kT} [when n tends to "∞" ] = 1 / (1  exp{ hν/kT})
N = N_{0} /(1  exp{ hν/kT})
When n tends to infinity E_{n} = nhν becomes infinite. There
is no confusion here because n is never infinite, However, The Boltzmann
formula has fixed the problem, and gives N = 0 in this case. In other
words, there is no oscillator inside the cavity that has an infinite
energy. We will note write : [i: 0 → "∞" ].
The expression of N_{0} = N(1  exp{ hν/kT}) gives us the
expression of N_{n}:
N_{n} = N_{0} exp{  nhν/kT} = N(1  exp{ hν/kT}) exp{  nhν/kT}
= N(1  exp{ hν/kT}) exp{  nhν/kT}
N_{n} = N(1  exp{ hν/kT}) exp{  nhν/kT}
The average energy E per oscillator, at the temperature T,
is then equal to:
E = ∑ N_{i} E_{i} /∑N_{i}
= (1/N)∑ N_{i} E_{i} = (1/N)∑ N(1  exp{ hν/kT}) exp{  ihν/kT} ihν
= hν(1  exp{ hν/kT})∑ i exp{  ihν/kT}
We have:
x = hν/kT
∑ i exp{  ix} =  d[∑exp{  ix}]/dx =  d[( 1  exp{ x(n+1)})/(1  exp{ x})]/dx
=  d[( 1  exp{ x(n+1)})/(1  exp{ x})]/dx . With n large, we have:
∑ i exp{  ix} =  d[1/(1  exp{ x})]/dx = exp{ x}/(1  exp{ x})^{2}
Finally:
E = hν(1  exp{ hν/kT}) exp{ hν/kT}/(1  exp{ hν/kT})^{2}
= hν exp{ hν/kT}/(1  exp{ hν/kT}) = hν /(exp{hν/kT}  1)
The average energy per oscillator:
E = hν /(exp{hν/kT}  1)
We have with RayleighJeans formula, set the number of oscillators dg
in the range ν and ν + dν, that is the number of stationary
states in the cavity which has, per volume unit, the folowing expression:
dg = (8π/c^{3}) ν^{2} dν.
The energy contained in this interval dν is the the average energy per
oscillator times the number of oscillators; that is:
dU = E dg = hν /(exp{hν/kT}  1) x (8π/c^{3}) ν^{2} dν
= (8πh/c^{3}) ν^{3} /(exp{hν/kT}  1) x dν
The spectral energy density is:
u(ν, T) = dU(ν, T)/dν = (8πh/c^{3}) ν^{3} /(exp{hν/kT}  1)
That is the Planck law
Transformed in the wavekength scale:
The spectral energy density is:
u(λ, T) = (8πhc/λ^{5}) x 1/(exp{hc/λkT}  1)
That is the Planck law.
The constant h remains unknown. Max Planck, in October 19, 1900, when he
presented his results to German Physics Society; his formula contained
two constants : C1 = 8πhc and C2 = hc/k. At this stage, it remains to
verify the formula, that is to determine the value of the constant h, set
this constant in the formula and see if this law is really the law of
the rays of blackbody.
