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(a)Use Equation5.113 to determine the energy density in the wavelength rangedλ. Hint: setÒÏ(Ó¬)dÓ¬=ÒÏ-(λ)»åλ, and solve forÒÏ(λ)-

(b)Derive the Wien displacement law for the wavelength at which the blackbody energy density is a maximum
λmax=2.90×10-3mKT

You'll need to solve the transcendental equation(5×x)=5e-x, using a calculator (or a computer); get the numerical answer accurate to three significant digits.

Short Answer

Expert verified

aÒÏ-λ=16Ï€2hcλ5e2hc/ke°Õλ-1bλmax=2.90×10-3mKT

Step by step solution

01

Given

The energy density in wavelength range is given by:

ÒÏÓ¬dÓ¬=ÒÏ-λ»åλ

02

Determining the energy density in the wavelength rangedλ

Ó¬=2Ï€c=2Ï€cλ,sodÓ¬=-2Ï€³¦Î»2»åλ,andÒÏÓ¬=hÏ€2c32Ï€³¦3λ3e2Ï€³ó³¦/kB°Õλ-1ÒÏÓ¬dÓ¬=8Ï€³ó1λ3e2Ï€³ó³¦/kB°Õλ-1-2Ï€³¦Î»2»åλ=ÒÏ-λ»åλÒÏλ=16Ï€2hcλ3e2Ï€³ó³¦/kB°Õλ-1-

For density we want only the size of the interval, not its sign)

03

Deriving the Wien displacement law for wavelength

To find maximize, we need to calculate:

dÒÏ-/»åλ=0;0=16Ï€2hc5λ6e2Ï€³ó³¦/kB°Õλ-1-e2Ï€³ó³¦/kB°Õλ2Ï€³ó³¦/kBTλ6e2Ï€³ó³¦/kB°Õλ-12-1λ2A=2Ï€³ó³¦kBT5=AλeA/γeA/γ-15e-A/λ=5-Aλ

Using mathematicAλ=4.966

λmax=A4.966=2πhc4.966kBTλmax=2.90×10-3mKT

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Most popular questions from this chapter

(a) If ψaandψb are orthogonal, and both normalized, what is the constant A in Equation 5.10?

(b) Ifrole="math" localid="1658225858808" ψa=ψb (and it is normalized), what is A ? (This case, of course, occurs only for bosons.)

We can extend the theory of a free electron gas (Section 5.3.1) to the relativistic domain by replacing the classical kinetic energy, E=p2/2m,,with the relativistic formula, E=p2c2+m2c4-mc2. Momentum is related to the wave vector in the usual way: p=hk. In particular, in the extreme relativistic limit, E≈pc=hck.

(a) Replace h2k2n Equation 5.55 by the ultra-relativistic expression, hck, and calculateEtotin this regime.

dE=h2k22mVÏ€2k2dk (5.55).

(b) Repeat parts (a) and (b) of Problem 5.35 for the ultra-relativistic electron gas. Notice that in this case there is no stable minimum, regardless of R; if the total energy is positive, degeneracy forces exceed gravitational forces, and the star will expand, whereas if the total is negative, gravitational forces win out, and the star will collapse. Find the critical number of nucleons, Nc , such that gravitational collapse occurs for N>N_{C}is called the Chandrasekhar limit.

(c) At extremely high density, inverse beta decaye-+p+→n+v,converts virtually all of the protons and electrons into neutrons (liberating neutrinos, which carry off energy, in the process). Eventually neutron degeneracy pressure stabilizes the collapse, just as electron degeneracy does for the white dwarf (see Problem 5.35). Calculate the radius of a neutron star with the mass of the sun. Also calculate the (neutron) Fermi energy, and compare it to the rest energy of a neutron. Is it reasonable to treat a neutron star non relativistic ally?

(a) Figure out the electron configurations (in the notation of Equation

5.33) for the first two rows of the Periodic Table (up to neon), and check your

results against Table 5.1.

1s22s22p2(5.33).

(b) Figure out the corresponding total angular momenta, in the notation of

Equation 5.34, for the first four elements. List all the possibilities for boron,

carbon, and nitrogen.

LJ2S+1 (5.34).

Check the equations 5.74, 5.75, and 5.77 for the example in section 5.4.1

(a) Calculate<1/r1-r2>for the stateψ0(Equation 5.30). Hint: Dod3r2integral

first, using spherical coordinates, and setting the polar axis alongr1, so

that

ψ0r1,r2=ψ100r1ψ100r2=8Ï€²¹3e-2r1+r2/a(5.30).

r1-r2=r12+r22-2r1r2³¦´Ç²õθ2.

Theθ2integral is easy, but be careful to take the positive root. You’ll have to

break ther2integral into two pieces, one ranging from 0 tor1,the other fromr1to∞

Answer: 5/4a.

(b) Use your result in (a) to estimate the electron interaction energy in the ground state of helium. Express your answer in electron volts, and add it toE0(Equation 5.31) to get a corrected estimate of the ground state energy. Compare the experimental value. (Of course, we’re still working with an approximate wave function, so don’t expect perfect agreement.)

E0=8-13.6eV=-109eV(5.31).

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