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Chapter 6: Q 6.8 (page 227)

The energy required to ionise a hydrogen atom is 13.6 eV, so you might expect that the number of ionised hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionised hydrogen is much larger, nearly one atom in 10,000. Explain why this result is not a contradiction, and why it would be incorrect to try to calculate the fraction of ionised hydrogen using the methods of this section.

Short Answer

Expert verified

It makes estimating the probabilities unknown, it does make the ionisation state more likely than using solely Boltzmann factors.

Step by step solution

01

Given information

The energy required to ionise a hydrogen atom is 13.6 eV, so you might expect that the number of ionised hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionised hydrogen is much larger, nearly one atom in 10,000.

02

Explanation

The probability of ionised atoms is equal to Boltzmann factors e-I/kTwhere I is the ionisation energy and T is the temperature, plus the Boltzmann factors multiplied by the degeneracy of the ionised state, which is essentially limitless. Although this makes estimating the probabilities unknown, it does make the ionisation state more likely than using solely Boltzmann factors. The likelihood does, in fact, depend on the number density of electrons in the surroundings.

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Most popular questions from this chapter

2. Consider a classical particle moving in a one-dimensional potential well u(x), as shown The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

where each integral is over the entire xaxis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(x)in a Taylor series about the equilibrium point x0: u(x)=ux0+x-x0dudxx0+12x-x02d2udx2x0

+13!x-x03d3udx3x0+鈰�

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x=x0.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for xbecome difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from zo by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series foru(x)

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

u(x)=u0x0x12-2x0x6

Sketch this function, and show that the minimum of the potential well is at x=x0, with depth u0. For argon, x0=3.9Aand u0=0.010eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u0. Evaluate the result numerically for argon, and compare to the measured value

伪=0.0007K-1(at80K)

For a diatomic gas near room temperature, the internal partition function is simply the rotational partition function computed in section 6.2, multiplied by the degeneracy Zeof the electronic ground state.

(a) Show that the entropy in this case is

S=NkInVZeZrotNvQ+72.

Calculate the entropy of a mole of oxygen at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book.

(b) Calculate the chemical potential of oxygen in earth's atmosphere near sea level, at room temperature. Express the answer in electron-volts

A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = m渭B, where 渭 the constant is 1.03 x 10-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T =-300 K.

The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation, defined as follows.

(a) For each atom in the five-atom toy model of Figure 6.5, compute the deviation of the energy from the average energy, that is, Ei-E炉,fori=1to5. Call these deviations 螖Ei.

(b) Compute the average of the squares of the five deviations, that is, 螖Ei2炉. Then compute the square root of this quantity, which is the root-mean- square (rms) deviation, or standard deviation. Call this number 蟽E. Does 蟽Egive a reasonable measure of how far the individual values tend to stray from the average?

(c) Prove in general that

蟽E2=E2炉-(E炉)2

that is, the standard deviation squared is the average of the squares minus the square of the average. This formula usually gives the easier way of computing a standard deviation.

(d) Check the preceding formula for the five-atom toy model of Figure 6.5.

Derive equation 6.92 and 6.93 for the entropy and chemical potential of an ideal gas.
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