Neutrons are released into an evacuated cubical chamber and lowered to a very cold T = 1 mK temperature. What is their mean height above the floor of the chamber, once their gravitational energy has reached an equilibrium distribution? (Show work! Derive a formula and quote a number.)

Solution:

- Concepts:

Boltzmann distribution - Reasoning:

We have a macroscopic-size chamber. - Details of the calculation:

The probability per unit length of finding the neutron at height z is

P(z) = P(0)exp(-mgz/(k_{B}T)).

1 = P(0)∫_{0}^{∞}exp(-mgz/(k_{B}T))dz = P(0)k_{B}T/(gm), P(0) = gm/(k_{B}T).

<z> = gm/(k_{B}T) ∫_{0}^{∞}z exp(-mgz/(k_{B}T)) dz = k_{B}T/(gm).

k_{B}= 1.380658 *10^{-23}J/K.

m_{n}= 1.6749286 * 10^{-27}kg.

<z> = 0.841 m.

Consider ideal gas particles
of mass m at temperature T. What is the average speed <v> (not the rms <v^{2}>^{½}
) of the particle in terms of m, T and the Boltzmann constant k_{B}?

∫_{0}^{∞}x^{2n }exp(-x^{2})dx = π^{½}
(1*3*5* ... *(2n -1))/2^{n+1}

Solution:

- Concepts:

Boltzmann statistics - Reasoning:

The Boltzmann distribution gives the probability of finding particles with energy E at a given temperature T. - Details of the calculation:

Boltzmann statistics: N_{i}/N = g_{i}exp(-E_{i}/kT)/∑_{i}(g_{i}exp(-E_{i}/kT))

Here g_{i}is the degeneracy.

For this problem:

n(v)dv/N = P(v)dv = g(v)exp(-mv^{2}/(2kT))dv/∫_{0}^{∞}g(v')exp(-mv'^{2}/(2kT))dv'

= v^{2}exp(-mv^{2}/(2kT))dv/∫_{0}^{∞}v'^{2}exp(-mv'^{2}/(2kT))dv',

since g(v) is proportional to v^{2}.

1/∫_{0}^{∞}v'^{2}exp(-mv'^{2}/(2kT))dv' = (m/(2kT))^{3/2}(4/π^{½}).

So <v> = ∫_{0}^{∞}vP(v)dv = (m/(2kT))^{3/2}(4/π^{½})_{ }∫_{0}^{∞}v^{3}exp(-mv^{2}/(2kT))dv

= (m/(2kT))^{3/2}(2/π^{½})_{ }∫_{0}^{∞}v^{2}exp(-mv^{2}/(2kT))dv^{2}= (8kT/(mπ))^{½}.

Some additional remarks:

The Boltzmann distribution describes the distribution of energy among classical (distinguishable) particles.

P(E) = A exp(-E/(kT))

It can be used to evaluate the average energy per particle in the circumstance where there is no energy-dependent density of states to skew the distribution.

To represent the probability for a given energy, it must be normalized to a probability of 1.

∫_{0}^{∞}Aexp(-E/(kT))dE = 1 ==> A = 1/(kT).

This normalized distribution function can then be used to evaluate the average energy.

<E> = (kT)^{-1}∫_{0}^{∞}Eexp(-E/(kT))dE = kT.

This is the average energy if the energy is randomly distributed among the available energy states and there is no energy-dependent density of states.

Note that this average energy for randomly distributed energy is not the same as the average kinetic energy.

The average kinetic energy is found by evaluating

<v^{2}> = ∫_{0}^{∞}v^{2}P(v)dv = (m/(2kT))^{3/2}(4/π^{½})_{ }∫_{0}^{∞}v^{4}exp(-mv^{2}/(2kT))dv = 3kT/m.

½m<v^{2}> = (3/2)kT.

We have an energy dependent density of states.

We can use E instead of v as the variable we integrate over.

g(v)dv = g(E)dE = g(E)mvdv, g(E) = g(v)/(mv).

Since g(v) is proportional to v^{2}, g(E) is proportional to E^{½}.

We can then rewrite all the integrals in terms of E instead of v.

Suppose a small meteorite makes a hole of area A = 1 mm^{2} in the wall
of a spaceship. The habitable volume of the spaceship is V = 10 m^{3}.
The temperature of the air in the spaceship is
T = 27^{o} C and the pressure P = 105 kPa. The molar mass of air is M =
29 g /mole. Estimate, how much time will be available to astronauts to put on
spacesuit, if the pressure should not drop by more than to 50% of its initial
value.

Maxwell-Boltzmann speed distribution: f(v) = (m/(2πkT))^{3/2 }4πv^{2}exp(-mv^{2}/(2kT)

Solution:

- Concepts:

The Maxwell Boltzmann distribution, ideal gas law - Reasoning:

We need to find the escape rate R_{esc}= -dN/dt, then we can integrate to find N(t). The rate with which the molecules escape from the hole

depends on their average speed. - Details of the calculation:

Consider an area dA and gas molecules with speeds between v and v + dv in a ring at a distance between r and r + dr from dA. Assume that the gas molecules travel in straight lines.

The probability that gas molecules from the ring will reach the area dA is

dA cosθ/(4πr^{2}). This is the solid angle subtended by dA at any dV in the ring.

The number of molecules from the ring reaching dA with speeds between v and v + dv is

nf(v)dv2πr^{2}sinθ dr dθ dA cosθ/(4πr^{2}) = nf(v)dv sinθ dr dθ dA cosθ/2.

Here n = N/V is the density of the gas.

The total number of molecules with speeds between v and v + dv reaching dA in a time interval dt comes from rings with radii between r = 0 and dr = vdt.

It is therefore given by

∫_{0}^{π/2}dθ nf(v)dv dA vdt sinθ cosθ/2 = ¼ n f(v) dv dA vdt.

Integrating over all speeds we find R_{esc}= ¼ n dA <v>,

where R_{esc}denotes rate at which gas molecules reach dA and <v> = ∫_{0}^{∞}vf(v) dv.

For the Maxwell Boltzmann speed distribution we find

<v> = (8kT/(mπ))^{½}, where m is the mass of a gas molecule.

R_{esc}= (P/kT) dA (kT/(2mπ))^{½}.R

_{esc}= -dN/dt is the rate at which molecules are escaping from the chamber.

dN/dt = -(N/V) dA (kT/(2mπ))^{½}.

N(t) = N_{0}exp(-Ct), where C = (dA/V) (kT/(2mπ))^{½}= (dA/V) (RT/(2Mπ))^{½}

If N/N_{0}= ½, then t = ln(2)/C.

C = (10^{-6}m^{2}/10 m^{3})*[(8.314 J/K mole * 300 K)/(2* 29*10^{-3}kg/mole * π)]^{½}_{ }= 1.71*10^{-3}/s.

t = 592 s ~ 10 minutes.