Subfield: Thermodynamics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: For black body radiation, Wien's Displacement Law states that its spectral radiance will peak at
$$
\lambda_{\text {peak }}=\frac{b}{T}
$$

where $b=2.89777 \times 10^{-3} \mathrm{mK}$, and $T$ is the temperature of the object. When QiLin tried to reproduce this in a lab, by working with a tungsten-filament lightbulb at $2800 \mathrm{~K}$, he computed a different value for $b$ by measuring the peak wavelength using a spectrometer and multiplying it with the temperature. He hypothesizes that this discrepancy is because tungsten is not an ideal black body. The graph below, courtesy of the CRC Handbook of Chemistry and Physics, shows the emissivity of tungsten at various conditions (the units for wavelength is $\mathrm{nm}$ ).



<img_4252>

Assuming QiLin's hypothesis is correct, and assuming there were no other errors in the experiment, how off was his value for $b$ ? Submit $\frac{\left|b_{\text {theory }}-b_{\text {experiment }}\right|}{b_{\text {theory }}}$ as a decimal number, to one significant digit (giving you room to estimate where the points are).

Solution:
Recall Planck's Law, which says the spectral radiance of a black body is given by
$$
B_{0}(\lambda, T)=\frac{2 h c^{3}}{\lambda^{5}} \frac{1}{\exp \left(\frac{h c}{\lambda k_{B} T}\right)-1}
$$

The regular Wien's Displacement Law can be derived by finding the peak by computing $\frac{\partial B_{0}}{\partial \lambda}$, to find the wavelength associated with the maximal radiance. For a nonideal body with emissivity $\epsilon(\lambda, T)$, we can write the radiance as

$$
B(\lambda, T)=B_{0}(\lambda, T) \epsilon(\lambda, T)
$$

We can estimate $\epsilon(\lambda, T)$ by looking at the given graph. The tungsten is at $2800 \mathrm{~K}$, so we will use the red line, and assuming it is near a black body, the peak wavelength should be around $1000 \mathrm{~nm}$. Performing a linear approximation around $1000 \mathrm{~nm}$, we get

$$
\epsilon(\lambda, T)=-173333\left(\lambda-1000 \cdot 10^{-9}\right)+0.366
$$

where $\lambda$ is in meters. Numerically finding the maximum of $B(\lambda, T)$ with respect to $\lambda$ (i.e. with a graphing calculator), we get the new peak wavelength to be $\lambda_{\text {new }}=949 \mathrm{~nm}$, while the old peak wavelength (assuming a perfect blackbody) is $\lambda_{\text {old }}=1035 \mathrm{~nm}$, and their percent difference (rounded to 1 significant digit) is 0.08

Final answer: 0.08