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Using the relationship between temperature and MandN given in (9-16) and that betweenE andn in (9-6), obtain equation (9-17) from (9- 12). The first sum given In Exercise 30 will be useful.

Short Answer

Expert verified

The expression for P(En) is NN+Me-ln(1+NM)

Step by step solution

01

Boltzmann Distribution.

The Boltzmann distribution for a system of particles in terms of n, M, and N is given by:

P(En)=NM+Ne-n鈥夆赌ln(1+NM)

Boltzmann Probability in terms ofEnandkBT:

P(En)=e-EnkBTn=0e-EnkBT

02

System of  N Harmonic Oscillators.

En=nh0P(En)=e-nh0kBTn=0e-nh0kBT=e-nh0kBTn=0[e-nh0kBT]n

03

Properties.

n=0xn=11-x

Denominator of P:

P(En)=e-nh0kBT11-e-nh0kBT=(1-e-nh0kBT)e-nh0kBTkBT=h0ln(1+NM)h0kBT=ln(1+NM)

On further calculation,

P(En)=(1-e-ln(1+NM))e-ln(1+NM)=1-1e-ln(1+NM)e-ln(1+NM)=1-11+NMe-ln(1+NM)=1+NM-11+NMe-ln(1+NM)=NM(M+N)Me-ln(1+NM)=NN+Me-ln(1+NM)

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Most popular questions from this chapter

The exact probabilities of equation (9-9) rest on the claim that the number of ways of addingNdistinct non-negative integer to give a total ofM is (M+N-1)![M!(N-1)!] . One way to prove it involves the following trick. It represents two ways that Ndistinct integers can add toM-9 and5, respectively. In this special case.

The X's represent the total of the integers, M-each row has 5. The 1'srepresent "dividers" between the distinct integers of which there will of course be N-Ieach row has8 . The first row says thatn1 is3 (three X'sbefore the divider between it andn2 ), n2is0 (noX's between its left divider withn1 and its right divider withn3 ),n3 ) is1 . n4throughn6 are0 , n7is1 , and n8and n9are0 . The second row says that n2is 2. n6is 1, n9is2 , and all othern are0 . Further rows could account for all possible ways that the integers can add toM . Argue that properly applied, the binomial coefficient (discussed in AppendixJ ) can be invoked to give the correct total number of ways for anyN andM .

For a room3.0mtall, by roughly what percent does the probability of an air molecule being found at the ceiling differ from that of an equal speed molecule being found at the floor? Ignore any variation in temperature from floor to ceiling.

Discusses the energy balance in a white dwarf. The tendency to contract due to gravitational attraction is balanced by a kind of incompressibility of the electrons due to the exclusion principle.

(a) Matter contains protons and neutrons, which are also fanions. Why do the electrons become a hindrance to compression before the protons and neutrons do?

(b) Stars several times our Sun's mass has sufficient gravitational potential energy to collapse further than a white dwarf; they can force essentially all their matter to become neutrons (formed when electrons and protons combine). When they cool off, an energy balance is reached like that in the white dwarf but with the neutrons filling the role of the incompressible fermions. The result is a neutron star. Repeat the process of Exercise 89. but assume a body consisting solely of neutrons. Show that the equilibrium radius is given by

R=322G(322mn8M)1/3

(c) Show that the radius of a neutron star whose mass is twice that of our Sun is only about10km .

What information would you need to specify the macro-state of the air in a room? What information would you need to specify the microstate?

According to Wien's law, the wavelengthmaxat which the thermal emission of electromagnetic energy from a body of temperatureTis maximum obeysmaxT=2.898103mK.Show that this law follows from equation (9-47). To do this. Usef=c/to expressin terms ofrather than f, then obtain an expression that, when solved, would yield the wavelength at which this function is maximum. The transcendental equation cannot be solved exactly, so it is enough to show that=(2.898103mK)/T solves it to a reasonable degree of precision.

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