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At very high temperatures (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the "nucleon." (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron's mass is higher than the proton's by 2.3 x 10-30 kg, its energy is higher by this amount times c2. Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 1011 K. What fraction of the nucleons at that time were protons, and what fraction were neutrons?

Short Answer

Expert verified

The fraction of neutron is 0.462 and

The fraction of neutron is 0.538.

Step by step solution

01

Given information

At very high temperatures (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the "nucleon." (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron's mass is higher than the proton's by 2.3 x 10-30 kg, its energy is higher by this amount times c2. Suppose, then, that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 1011 K.

02

Explanation

The neutron and the proton are two states of the nucleon in the early cosmos, and the energies of the proton and neutron are:

Ep=mpc2En=mnc2

The probabilities of two states are:

Pp=1Ze-Ep/kTPn=1Ze-En/kTPp=1Ze-mpc2/kTPn=1Ze-mnc2/kT

The ratio of the probabilities is:

PnPp=e-mnc2/kTe-mpc2/kT=e-Δmc2/kT

Where mis the mass difference between neutron and proton.

PnPp=e-2.3×10-30kg3.0×108m/s2/1.38×10-23J/K1011K

PnPp=0.86

This means that for every 100 protons, there are 86 neutrons; the total number is 100+86 =186; the neutron fraction is:

fn=86100+86=0.462fn=0.462

Fraction of proton is:

fp=100100+86=0.538fp=0.538

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Most popular questions from this chapter

For a mole nitrogen (N2)gas at room temperature and atmospheric pressure, compute the following U,H,F,G,Sand μ. (The electronic ground state of nitrogen is not degenerate.)

Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule at height z.)

Estimate the partition function for the hypothetical system represented in Figure 6.3. Then estimate the probability of this system being in its ground state.

This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on. (

a) Prove by long division that

11-x=1+x+x2+x3+

For what values of x does this series have a finite sum?

(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part (a) to simplify your answer as much as possible.

(c) Use formula 6.25 to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.

(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 3.25.

(e) If you haven't already done so in Problem 3.25, compute the heat capacity of this system and check t hat it has the expected limits as T0 and T.

In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula

Enϵ1.03n-0.03n2,n=0,1,2,

where ϵ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n 15.) Use a computer to calculate the partition function, average energy, and heat capacity of a system with this set of energy levels. Include all levels through n = 15, but check to see how the results change when you include fewer levels Plot the heat capacity as a function of kT/ϵ. Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in Figure 1.13.

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