Projectile's charge Z Projectile's E (MeV)

Contents

The nucleus

Statistical Mechanics

Related topics

# 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 (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 = En = 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, N0 the number of oscillators (with the same frequency ν) with 0(zero) quantum, or 0 energy, ..., and Nn the number of oscillators (always with the same frequency ν) with n quanta, or nhν energy, we have: N = N0 + N1 + ... + Nn = ∑ Ni [ 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 Ni of the different levels "i" of energy ihν is set by the Boltzmann formula:

Pi = Probability to have Ni oscillators in the state "i", then with energy Ei = ihν Ni = N x Pi = exp{ - Ei/kT}/Z
where Z is the partition function for the oscillators = ∑ exp{ - Ei/kT} [i: 0 → ∞ ]

Similarly, we have:
P0 = Probability to have N0 oscillators in the state "0" or ground state and N0 = N x P0

We can then write the ratio: Ni/N0 = exp{ - (Ei - E0)/kT}= exp{ - (ihν - 0)/kT} and we get the:
Boltzmann formula: Nn/N0 = exp{ - nhν/kT}

The total number N is calculated as:
N = ∑ Ni [ i: 0 → "∞" ] = N0∑ exp{ - nhν/kT} = N0∑ 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 = N0 /(1 - exp{- hν/kT})

When n tends to infinity En = 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 N0 = N(1 - exp{- hν/kT}) gives us the expression of Nn: Nn = N0 exp{ - nhν/kT} = N(1 - exp{- hν/kT}) exp{ - nhν/kT} = N(1 - exp{- hν/kT}) exp{ - nhν/kT}

Nn = N(1 - exp{- hν/kT}) exp{ - nhν/kT}

The average energy E per oscillator, at the temperature T, is then equal to:
E = ∑ Ni Ei /∑Ni = (1/N)∑ Ni Ei = (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 Rayleigh-Jeans 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/c3) ν2 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/c3) ν2 dν = (8πh/c3) ν3 /(exp{hν/kT} - 1) x dν

The spectral energy density is: u(ν, T) = dU(ν, T)/dν = (8πh/c3) ν3 /(exp{hν/kT} - 1) That is the Planck law

Transformed in the wavekength scale:
The spectral energy density is:
u(λ, T) = (8πhc/λ5) 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. Web ScientificSentence