Lecture 8 Solutions: Blackbody Radiation
Practice Problem Solutions
Student note: These are model solutions written to show every step, units, and checks. I do not expect your work to be this detailed; shorter solutions are fine if your setup, units, and reasoning are correct.
HW note: Homework uses only a subset of these problems; the full set is included for future study and self checking.
Core Problems
Problem 1: Wien’s Law (Temperature from Peak)
Restatement: Calculate the surface temperature of a star whose spectrum peaks at 580 nm.
Given:
- Peak wavelength: \(\lambda_{\text{peak}} = 580\) nm
- Wien’s Law constant: \(b = 2.9 \times 10^6\) nm·K
Find: Surface temperature \(T\) in Kelvin.
Equation: \[T = \frac{b}{\lambda_{\text{peak}}} = \frac{2.9 \times 10^6 \text{ nm·K}}{\lambda_{\text{peak}}}\]
Solution:
\[T = \frac{2.9 \times 10^6 \text{ nm·K}}{580 \text{ nm}} = 5000 \text{ K}\]
Unit check: \(\frac{\text{nm·K}}{\text{nm}} = \text{K}\) ✓
Sanity check: The Sun’s surface temperature is ~5800 K and it peaks at ~500 nm. A peak at 580 nm (longer wavelength, slightly redder) should correspond to a slightly cooler star — and 5000 K is indeed slightly cooler than the Sun. ✓
Answer: \(T = 5000\) K. This star is slightly cooler than the Sun.
What you should learn from this: Wien’s Law is a one-step thermometer: divide the constant by the peak wavelength. Always check that both are in the same wavelength units (here, nm).
Problem 2: Wien’s Law Reverse (Peak from Temperature)
Restatement: Find the peak wavelength for a 4000 K star and identify the spectral region.
Given:
- Surface temperature: \(T = 4000\) K
- Wien’s Law constant: \(b = 2.9 \times 10^6\) nm·K
Find: Peak wavelength \(\lambda_{\text{peak}}\) in nm; spectral region.
Equation: \[\lambda_{\text{peak}} = \frac{b}{T} = \frac{2.9 \times 10^6 \text{ nm·K}}{T}\]
Solution:
\[\lambda_{\text{peak}} = \frac{2.9 \times 10^6 \text{ nm·K}}{4000 \text{ K}} = 725 \text{ nm}\]
Unit check: \(\frac{\text{nm·K}}{\text{K}} = \text{nm}\) ✓
Spectral region: Visible light spans roughly 400–700 nm. At 725 nm, this star peaks in the near-infrared, just beyond the red end of visible light. It would appear reddish to our eyes because the visible portion of its emission is dominated by red wavelengths.
Sanity check: A cooler star (4000 K < 5800 K for the Sun) should peak at a longer wavelength — and 725 nm > 500 nm (Sun’s peak). Cooler = redder. ✓
Answer: \(\lambda_{\text{peak}} = 725\) nm (near-infrared). The star would appear reddish.
What you should learn from this: Wien’s Law works both ways — temperature → wavelength or wavelength → temperature. A 4000 K star peaks just outside visible light, explaining why cool stars appear red.
Problem 3: Stefan-Boltzmann (Luminosity Ratio)
Restatement: Star A is 3× hotter than Star B with the same radius. How much more luminous is Star A?
Given:
- \(T_A = 3 T_B\)
- \(R_A = R_B\)
Find: Luminosity ratio \(L_A / L_B\).
Equation: \[L = 4\pi R^2 \sigma T^4 \quad \Rightarrow \quad \frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4\]
Solution:
\[\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4 = (1)^2 \times (3)^4 = 1 \times 81 = 81\]
Unit check: Ratio of luminosities — dimensionless. ✓
Sanity check: Temperature appears to the fourth power, so even a modest temperature increase has a dramatic effect on luminosity. Tripling the temperature → \(3^4 = 81\)× more luminous. This extreme sensitivity is why temperature matters more than size for stellar luminosity. ✓
Answer: Star A is 81 times more luminous than Star B.
What you should learn from this: The \(T^4\) in Stefan-Boltzmann makes temperature the dominant factor in luminosity. A factor of 3 in temperature produces a factor of 81 in luminosity — temperature matters far more than size.
Problem 4: L-T-R Relationship (Radius Ratio)
Restatement: Two stars have the same temperature but one is 400× more luminous. Compare their radii.
Given:
- \(T_A = T_B\)
- \(L_A = 400 \, L_B\)
Find: Radius ratio \(R_A / R_B\).
Equation: \[\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4\]
Solution:
Since \(T_A = T_B\), the temperature ratio is 1:
\[400 = \left(\frac{R_A}{R_B}\right)^2 \times (1)^4 = \left(\frac{R_A}{R_B}\right)^2\]
\[\frac{R_A}{R_B} = \sqrt{400} = 20\]
Unit check: Ratio of radii — dimensionless. ✓
Sanity check: At the same temperature, the only way to be more luminous is to be bigger (more surface area emitting light). \(400\times\) more luminous → \(\sqrt{400} = 20\times\) larger radius. This is physically reasonable — giant stars can be 10–100× the Sun’s radius. ✓
Answer: The brighter star has 20 times the radius of the fainter star.
What you should learn from this: At fixed temperature, \(L \propto R^2\). The square root relationship means you need a large luminosity difference (400×) to imply a large radius difference (20×). This is how astronomers identify giant and supergiant stars — they’re far more luminous than main-sequence stars of the same color.
Problem 5: Why the Sun Isn’t Green
Restatement: Why does the Sun appear white, not green, even though its peak wavelength (~500 nm) is in the green?
Key insight: The blackbody spectrum is a broad curve, not a narrow spike.
Answer: The Sun’s spectrum peaks near 500 nm (green), but this does not mean the Sun appears green. Here’s why:
The blackbody curve is broad. The Sun emits strongly across the entire visible range — violet through red. The peak at 500 nm is a gentle maximum, not a sharp spike. There is nearly as much blue and red light as green.
Our eyes integrate. Human vision combines all incoming visible wavelengths. When you mix roughly equal amounts of all visible colors, the result is white. (This is how white LEDs work — they combine wavelengths across the visible spectrum.)
If the Sun emitted only at 500 nm (like a laser), it would indeed appear green. But a blackbody emits at all wavelengths, with a smooth distribution.
Common misconception: “The Sun is yellow.” From space, the Sun appears white. It looks yellowish from Earth’s surface because Rayleigh scattering removes some blue light from the direct beam (L7 connection).
What you should learn from this: “Peak wavelength” does not mean “color.” Color perception depends on the entire spectrum, not just the peak. This distinction matters whenever you use Wien’s Law — it tells you where the peak is, not what color your eyes see.
Problem 6: Star Color Ranking
Restatement: Rank a red, blue, and yellow star from coolest to hottest.
Answer: Coolest to hottest: red → yellow → blue.
Reasoning from Wien’s Law: \(\lambda_{\text{peak}} = b/T\), so hotter stars peak at shorter wavelengths.
- Red star: peaks at long wavelengths (\(\sim 700\)+ nm) → lowest temperature (\(\sim 3000\)–\(4000\) K)
- Yellow star: peaks at intermediate wavelengths (\(\sim 500\)–\(600\) nm) → moderate temperature (\(\sim 5000\)–\(6000\) K, like the Sun)
- Blue star: peaks at short wavelengths (\(\sim 300\)–\(450\) nm) → highest temperature (\(\sim 10{,}000\)–\(30{,}000\) K)
Common misconception: Everyday experience associates blue with cold (water faucets, ice). In astrophysics, the opposite is true: blue = hot, red = cool. This is a physics result from Wien’s Law.
What you should learn from this: Star color is a thermometer. The sequence red → orange → yellow → white → blue corresponds to increasing surface temperature. This will be essential for understanding the Hertzsprung-Russell diagram later in the course.
Challenge Problems
Problem 7: Combined Wien + Stefan-Boltzmann
Restatement: Star X has \(T_X = 2T_Y\) and \(R_X = R_Y/2\). Compare their peak wavelengths and luminosities.
Given:
- \(T_X = 2T_Y\)
- \(R_X = R_Y / 2\)
Find: (a) Peak wavelength ratio; (b) Luminosity ratio.
Solution:
(a) Wien’s Law: \[\frac{\lambda_{\text{peak},X}}{\lambda_{\text{peak},Y}} = \frac{b/T_X}{b/T_Y} = \frac{T_Y}{T_X} = \frac{T_Y}{2T_Y} = \frac{1}{2}\]
Star X peaks at half the wavelength of Star Y (shorter wavelength = bluer).
Unit check: Ratio of wavelengths — dimensionless. ✓
(b) Stefan-Boltzmann Law: \[\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{1}{2}\right)^2 \times (2)^4 = \frac{1}{4} \times 16 = 4\]
Unit check: Ratio of luminosities — dimensionless. ✓
Sanity check: Star X is hotter (which increases luminosity as \(T^4\)) but smaller (which decreases luminosity as \(R^2\)). The \(T^4\) wins: \(2^4 = 16\) vs. \((1/2)^2 = 1/4\), and \(16 \times 1/4 = 4\). Temperature dominance confirmed. ✓
Answer: Star X peaks at half the wavelength of Star Y, and is 4× more luminous.
What you should learn from this: This problem shows the competition between size and temperature. Even though Star X is half the radius, its double temperature makes it 4× more luminous overall. The \(T^4\) dependence almost always wins.
Problem 8: Earth as a Blackbody
Restatement: Find Earth’s thermal emission peak wavelength given its effective temperature.
Given:
- Earth’s effective temperature: \(T = 255\) K
- Wien’s Law constant: \(b = 2.9 \times 10^6\) nm·K
Find: Peak wavelength \(\lambda_{\text{peak}}\) and spectral region.
Equation: \[\lambda_{\text{peak}} = \frac{b}{T}\]
Solution:
\[\lambda_{\text{peak}} = \frac{2.9 \times 10^6 \text{ nm·K}}{255 \text{ K}} = 1.14 \times 10^4 \text{ nm} = 11.4 \text{ μm}\]
Unit check: \(\frac{\text{nm·K}}{\text{K}} = \text{nm}\) ✓ ; conversion: \(1.14 \times 10^4 \text{ nm} \times \frac{1 \text{ μm}}{1000 \text{ nm}} = 11.4 \text{ μm}\) ✓
Spectral region: This is in the thermal infrared — far beyond visible light (400–700 nm). Earth’s thermal emission is invisible to human eyes.
Sanity check: Earth is much cooler than stars (255 K vs. thousands of K), so it should peak at much longer wavelengths than visible light. 11.4 μm is well into the infrared, as expected. ✓
Answer: \(\lambda_{\text{peak}} \approx 11.4\) μm (thermal infrared).
Why this matters: Greenhouse gases (CO₂, H₂O, CH₄) absorb strongly in the infrared. Earth’s thermal emission peaks right where these gases absorb, which is why the greenhouse effect works — it intercepts Earth’s outgoing thermal radiation. If Earth emitted in visible light, greenhouse gases wouldn’t matter.
What you should learn from this: Wien’s Law works for any thermal emitter, not just stars. Earth, planets, and even human bodies emit thermal radiation — all peaking in the infrared for objects at hundreds of Kelvin.
Problem 9: Size from L and T (Betelgeuse)
Restatement: Estimate \(R_{\text{Betelgeuse}}/R_\odot\) from its luminosity and temperature.
Given:
- Betelgeuse: \(T_B = 3500\) K, \(L_B = 10^5 \, L_\odot\)
- Sun: \(T_\odot = 5800\) K, \(L_\odot = 1 \, L_\odot\)
Find: Radius ratio \(R_B / R_\odot\).
Equation: From the L-T-R relationship in ratio form: \[\frac{L_B}{L_\odot} = \left(\frac{R_B}{R_\odot}\right)^2 \left(\frac{T_B}{T_\odot}\right)^4\]
Solve for the radius ratio: \[\frac{R_B}{R_\odot} = \left(\frac{L_B}{L_\odot}\right)^{1/2} \times \left(\frac{T_\odot}{T_B}\right)^2\]
Solution:
Step 1 — Calculate the temperature ratio: \[\frac{T_B}{T_\odot} = \frac{3500 \text{ K}}{5800 \text{ K}} = 0.603\]
Step 2 — Calculate the temperature factor (to the 4th power): \[\left(\frac{T_B}{T_\odot}\right)^4 = (0.603)^4 = 0.132\]
Step 3 — Solve for the radius ratio: \[10^5 = \left(\frac{R_B}{R_\odot}\right)^2 \times 0.132\]
\[\left(\frac{R_B}{R_\odot}\right)^2 = \frac{10^5}{0.132} = 7.58 \times 10^5\]
\[\frac{R_B}{R_\odot} = \sqrt{7.58 \times 10^5} = 871 \approx 880\]
Unit check: All ratios are dimensionless. ✓
Sanity check: Betelgeuse is a red supergiant — it should be enormous. At 880 \(R_\odot\), its radius would be: \[880 \times 6.96 \times 10^8 \text{ m} = 6.1 \times 10^{11} \text{ m} \approx 4.1 \text{ AU}\]
This means Betelgeuse would extend past the orbit of Mars (1.5 AU) and into the asteroid belt. This is consistent with the known properties of red supergiants. ✓
Answer: \(R_{\text{Betelgeuse}} \approx 880 \, R_\odot\) (~4 AU). Betelgeuse is large enough to swallow Mercury, Venus, Earth, Mars, and the entire asteroid belt.
What you should learn from this: The ratio method avoids needing \(\sigma\) or absolute values — you only need the luminosity ratio and the temperature ratio. This is how astronomers determine stellar radii without physically measuring them.
Problem 10: Why Blackbodies Matter
Restatement: Explain the utility of blackbody physics for astronomy and what additional information requires spectral lines.
Answer:
What blackbody physics provides:
Blackbody radiation is a universal property of thermal emitters. For any star (which approximates a blackbody), we can determine:
- Temperature — from Wien’s Law (\(\lambda_{\text{peak}} = b/T\)): measure where the spectrum peaks → get surface temperature
- Luminosity — from the Stefan-Boltzmann Law (\(L = 4\pi R^2 \sigma T^4\)): if you know \(T\) and \(R\), you know total power output
- Radius — from the L-T-R relationship: if you know \(L\) and \(T\), solve for \(R\)
These three properties (temperature, luminosity, radius) are fundamental characteristics of any star, and they come entirely from the smooth, continuous blackbody spectrum.
What requires spectral lines:
Blackbody spectra are smooth and featureless — they look the same for any object at a given temperature, regardless of composition. To determine:
- Composition — which elements are present in the atmosphere (L9: each element produces unique absorption/emission lines)
- Radial velocity — whether the star is approaching or receding (L10: Doppler shifts of spectral lines)
- Rotation rate — how fast the star spins (line broadening)
- Magnetic fields — field strength (Zeeman splitting of lines)
…you need the discrete features (spectral lines) superimposed on the smooth blackbody curve. Blackbody physics gives you the “big picture” (T, L, R); spectral lines give you the details (composition, motion, environment).
What you should learn from this: Blackbody radiation and spectral lines are complementary tools. The smooth curve tells you about thermal properties; the lines tell you about atomic physics and motion. Together, they give a remarkably complete picture of a distant object.