1. Cosmic Microwave Background
The CMB is a perfect blackbody with a temperature of ~3000 K. On the Earth, we see that the CMB has a blackbody temperature of ± K.

2. Cosmic Microwave Background
The CMB is a perfect blackbody with a temperature of ~3000 K. On the Earth, we see that the CMB has a blackbody temperature of ± K.
Using Wien’s displacement law:

3. Cosmic Microwave Background
The CMB is a perfect blackbody with a temperature of ~3000 K. On the Earth, we see that the CMB has a blackbody temperature of ± K.
Using Wien’s displacement law, For T = 3000 K, λpeak = × 10-3/3000 = μm.

4. Cosmic Microwave Background
The CMB is a perfect blackbody with a temperature of ~3000 K. On the Earth, we see that the CMB has a blackbody temperature of ± K.
Using Wien’s displacement law, For T = 3000 K, λpeak = × 10-3/3000 = μm For T = K, λpeak = × 10-3/ = mm.

5. Cosmic Microwave Background
The CMB is a perfect blackbody with a temperature of ~3000 K. On the Earth, we see that the CMB has a blackbody temperature of ± K.
Using Wien’s displacement law, For T = 3000 K, λpeak = × 10-3/3000 = μm For T = K, λpeak = × 10-3/ = mm.
Redshift,
= (1.097 × 10-3 − × 10-6)/(0.966 × 10-6) = 1135

6. Space Motion 2.
Barnard’s star is a M4V star with an apparent magnitude of It is the next closest star to the Earth after the α Centauri system (distance 1.34 pc), and has the largest known proper motion.

7. Space Motion 2.
From the equation for the Doppler effect for vr << c,
so that vr = c Δλ/λrest = ×105 ( – )/( ) = –112.8 km/s What does the –ve sign imply? Component of stellar velocity along the line of sight is towards the Sun. (Note: above measurements have been corrected for Earth’s rotation and orbit around the Sun, and are therefore centered on the Sun. Earth’s rotation speed is ~460 m/s at the equator, and orbital speed ~30 km/s.)

8. Space Motion 2.
Distance d = 1/p pc = 1/ = 1.82 pc

9. Space Motion 2. Distance d = 1/p pc = 1/0.54901 = 1.82 pc
Transverse velocity
vt = dμ = 1.82 × 3.086× π/(60×60×180)
= 2.82×1012 m/yr
= 89.4 km/s

10. Space Motion 2. Space velocity
v = √(vr2 + vt2) = √[(–112.8 ) ] =143.9 km/s

11. Work Function 3.
Kinetic Energy K = Ephoton − ϕ so that ϕ = Ephoton − K = hc/λ − K = (6.626× ×108)/100×10-9 − (5 × 1.602×10-19) = ×10-18 J = 7.4 eV

12. Compton Wavelength 4. For a proton with mass mp
  Δλ = (h/mpc) (1 – cos θ)
For θ = 0o, Δλ = 0; i.e., no change. Cannot impart forward momentum. Why? For θ = 90o, Δλ = h/mpc For θ = 180o, Δλ = 2h/mpc; i.e., largest change in wavelength. Recoils backwards.

13. Compton Wavelength 4. For a proton with mass mp
  Δλ = (h/mpc) (1 – cos θ)
For θ = 0o, Δλ = 0; i.e., no change in wavelength For θ = 90o, Δλ = h/mpc For θ = 180o, Δλ = 2h/mpc; i.e., largest change in wavelength.

14. Compton Wavelength 4. = 1.32×10-15 m By comparison, for the electron
For a proton with mass mp
  Δλ = (h/mpc) (1 – cos θ)
For θ = 0o, Δλ = 0; i.e., no change For θ = 90o, Δλ = h/mpc For θ = 180o, Δλ = 2h/mpc; i.e., largest change in wavelength.
Characteristic change in wavelength is for θ ≈ 90o,
  Δλ = h/mpc = (6.626×10-34)/(1.673× ×108)
= 1.32×10-15 m
By comparison, for the electron
Δλ = h/mec = (6.626×10-34)/(9.109× ×108)
= 2.43×10-12 m

15. Bohr’s Model for Hydrogenic Ions5.
For a proton with mass mp
  Δλ = (h/mpc) (1 – cos θ)

16. Bohr’s Model for Hydrogenic Ions5.
From Coulomb’s law
  F = q1 q2 / 4 π εo r = - Z e2 / 4 π εo r2 (-ve because Coulomb force directed inwards)
Using the reduced-mass system for the atom
- Z e2 / 4 π εo r2 = -μ v2/r (-ve because centripetal force directed inwards) so that kinetic energy of the atom K = ½ μ v = Z e2 / 8 π εo r.
Assuming that angular momentum is quantized such that μvr = nħ ½ μ v2 = ½ (μvr)2/μr = ½ (nħ)2/μr Solving for r r = n2 (4 π εo ħ2)/(μ Z e2 ) = (n2/Z) nm

17. Bohr’s Model for Hydrogenic Ions5.
Potential energy of the atom
  U = q1 q2 / 4 π εo r = - Z e2 / 4 π εo r Total energy of the atom E = K + U = Z e2 / 8 π εo r - Z e2 / 4 π εo r = -Z e2 / 8 π εo r Substitute for r E = - μ Z2 e4 / (32 π2 εo2 ħ2 n2) = (Z2/n2) eV

18. Bohr’s Model for Hydrogenic Ions5.
For Li III, Z = 3. Ground-star orbital radius, n = 1, and therefore
  r = (n2/Z) nm
= (12/3) nm = nm Ground-state energy E = (Z2/n2) eV
= (32/12) eV = eV
Ionization energy of Li III is therefore eV.

19. Electronic Transitions6.
Energy of emitted photon
Wavelength of emitted photon Possible transitions are 32, E = 1.9 eV, λ = 656 nm (Hα) 1, E = 12.1 eV, λ = 103 nm (Lyβ) 1, E = 10.2 eV, λ = 122 nm (Lyα) RH RH

20. Electronic Transitions7.
Wavelength of emitted photon
Lyman limit is 1, λ = 91 nm in the ultraviolet Balmer limit is 2, λ = 365 nm in the ultraviolet Paschen limit is 3, λ = 802 nm in the infrared RH

21. Excitation State of Interstellar Hydrogen
8. The fact that we can see stars and galaxies indicates that the interstellar medium is almost completely transparent at optical wavelengths (along lines of sight away from the Galactic plane). Explain why this implies that the bulk of hydrogen atoms cannot be at the n = 3 or higher levels. Do you also expect the interstellar medium to be transparent in the ultraviolet? Provide explanations for your answer Lyman limit is 1, λ = 91 nm in the ultraviolet Balmer limit is 2, λ = 365 nm in the ultraviolet Paschen limit is 3, λ = 802 nm in the infrared
If the majority of atoms are in the n ≥ 3, will get limit absorption in the Paschen and higher series at infrared at shorter wavelengths, thus producing strong absorption in the optical.
Hydrogen atoms at the n=1 or n=2 levels will absorb strongly in the ultraviolet.

22. de Broglie wavelength 9.
The de Broglie wavelength λ = h/p = h/μv Circumference of orbit = 2πr = nλ for constructive interference, so that πr = nλ = nh/μv i.e., μvr = nh/2π = nħ, which is Bohr’s condition for the quantization of angular momentum.

23. Degenerate Electron Pressure
10.
Heisenberg’s uncertainty principle Δx Δp ≈ ħ Δp ≈ ħ/Δx me Δv ≈ ħ/Δx
Δv ≈ ħ/(me Δx) ≈ 0.26 c

24. Sirius
Sirius is the brightest star in the night sky.

25. Sirius Bessel made the first parallax measurement of a star, 61 Cygni.
Suspecting that Sirius is close, Bessel then measured a parallax to Sirius of 0˶.379, corresponding to a distance of 2.64 pc.

26. Sirius
Bessel found that the apparent motion of Sirius across the sky exhibited a “wobble,” leading him to suggest the presence (and predict the position) of a companion star with an orbital period of ~50 years.

27. Sirius
In 1862, 26 years after the death of Bessel, the companion star was discovered at its predicted location.

28. Sirius
In 1915, Walter Adams showed that Sirius B has virtually the same spectral type (and therefore comparable surface temperature) as Sirius A, both early-A.

29. Sirius
In 1915, Walter Adams showed that Sirius B has virtually the same spectral type (and therefore comparable surface temperature) as Sirius A, both early-A.
This implies that Sirius B must be much smaller than Sirius A, somewhat smaller than the Earth!
From the orbits of the two stars about their center of mass, the masses of the two stars were deduced to be ~2.3 M for Sirius A and ~1 M for Sirius B.

30. Sirius
In 1915, Walter Adams showed that Sirius B has virtually the same spectral type (and therefore comparable surface temperature) as Sirius A, both early-A.
This implies that Sirius B must be much smaller than Sirius A, somewhat smaller than the Earth!
From the orbits of the two stars about their center of mass, the masses of the two stars were deduced to be ~2.3 M for Sirius A and ~1 M for Sirius B.
Today, we know Sirius B to be a white dwarf, a star with an original mass of ~5 M and which expelled most of its mass during its post-main-sequence evolution.

31. Sirius
The pressure (~107 times the pressure at the center of the Sun) and temperature (similar to the temperature at the center of the Sun) at the center of Sirius B is so high that that, if Sirius B shines by thermonuclear fusion, it would be several orders of magnitudes more luminous than observed.
In fact, hydrogen cannot be present at any appreciable amounts even just below the stellar surface.
Rather, the interior of Sirius B comprises (full ionized) atomic nuclei and free electrons.
In normal stars, the outward pressure resisting gravity is provided mainly by thermal gas pressure.
In white dwarfs, the outward pressure resisting gravity is provided mainly by electrons behaving according to Pauli’s exclusion principle.

32. Quantum States of Fermion Gas
Imagine fermions, in this case electrons, confined in a box. Thinking of electrons as being standing waves in a box, their wavelengths in each dimension are given by
where Nx, Ny, and Nz are integer quantum numbers associated with each direction.

33. Quantum States of Fermion Gas
Imagine fermions, in this case electrons, confined in a box. Thinking of electrons as being standing waves in a box, their wavelengths in each dimension are given by
where Nx, Ny, and Nz are integer quantum numbers associated with each direction.
Recall that the de Broglie wavelength is related to momentum by
The kinetic energy of a particle is therefore

34. Quantum States of Fermion Gas
Recall that Pauli’s exclusion principle states that no two electrons can have the same set of quantum numbers. Thus, at most, only two electrons (opposite spins) can occupy each state with the same quantum number Nx, Ny, and Nz.
At T = 0 K, electrons will fill all the lowest energy states, and the gas is complete degenerate. Even at T = 0 K, most of the particles in a degenerate fermion gas has kinetic energy.
The interior of a white dwarf is close to being a degenerate fermion gas.
The kinetic energy of this fermion gas provides pressure against gravity.

35. Natural Linewidth
11.
In the non-relativistic approximation, the energy of quantum state N (see Chap. 16)

36. Natural Linewidth
11.
Heisenberg’s uncertainty principle ΔE Δt ≈ ħ ΔE ≈ ħ/Δt ≈ ×10-34 / (2 π 10-8) ≈ ×10-34 / (2 π 10-8)
≈ 1.05×10-26 J ≈ 6.58×10-8 eV

37. Natural Linewidth
11.
Energy of a photon E = hc/λ Differentiating with respect to λ E/λ = hc (1/λ)/λ = hc (-1/λ2)
i.e., λ = -λ2 E / h c
= -(121×10-9)2 1.05×10-26 / 6.626× × = 7.74×10-16 m
= 7.74×10-7 nm In In the ground state, Δt =  and so ΔE = 0. Note that the width of natural broadening for the 21 line is λ/λ = 7.74×10-7/121 = 6.4×10-9; i.e., 6.4 parts of a billionth of the line wavelength.

38. Quantum States Hydrogen
12. n l ml ms

39. Quantum States Hydrogen
12. n 3 l 1 2 ml -1 -2 ms +½ -½ n l ml ms

40. Quantum States Hydrogen
12. n l ml ms
Number of degenerate levels at n = 1 is 2.
Number of degenerate levels at n = 2 is 8.

41. Redshift and Time Dilation
11. n 3 l 1 2 ml -1 - -2 ms +½ -½ n l ml ms
Number of degenerate levels at n = 3 is 18. In general, number of degenerate levels is 2n2.

42. Zeeman Effect
13.
Recall that Zeeman effect is splitting of the energy levels of electrons with different quantum numbers ml

43. Zeeman Effect 13. Three line components are at frequencies
where e is the electronic charge, B the field strength, and μ the reduced mass of the electron. Thus, frequencies of the three line components are ×1014 Hz, ×1014 Hz (Hα), and ×1014 Hz. Corresponding wavelengths are nm, nm (Hα), and nm.