1.88: Planck's Radiation Equation Fit to Experimental Data - Another Algorithm
- Page ID
- 157656
\[
\mathrm{n} :=42 \qquad \mathrm{i} :=1 . . \mathrm{n}
\nonumber \]
\(\rho_{i} :=\) | \(\lambda_{i} :=\) |
0.07 | 0.667 |
0.096 | 00.720 |
0.10 | 0.737 |
0.190 | 0.811 |
0.210 | 0.383 |
0.398 | 0.917 |
0.420 | 0.917 |
0.680 | 1.027 |
0.708 | 1.021 |
1.036 | 1.167 |
1.062 | 1.172 |
1.258 | 1.247 |
1.669 | 1.484 |
1.770 | 1.697 |
1.776 | 1.831 |
1.730 | 2.039 |
1.685 | 2.170 |
1.640 | 2.275 |
1.551 | 2.406 |
1.392 | 2.563 |
1.145 | 2.27 |
1.115 | 2.824 |
1.071 | 2.916 |
1.042 | 2.921 |
0.974 | 3.050 |
0.918 | 2.151 |
0.797 | 3.344 |
0.760 | 3.450 |
0.742 | 3.556 |
0.698 | 3.661 |
0.667 | 3.754 |
0.570 | 4.027 |
0.426 | 4.427 |
0.378 | 4.613 |
0.345 | 4.805 |
0.310 | 4.968 |
0.280 | 5.128 |
0.250 | 5.296 |
0.220 | 5.469 |
0.205 | 5.632 |
0.175 | 5.783 |
0.155 | 6.168 |
The data for this exercise is taken from page 19 of Eisberg and Resnick, Quantum Physics.
The values of rho are given in units of 103 joules/m3 and the values of lambda are given in 10‐6 m. The temperature is 1595 K.
Define Planck radiation function and first derivatives with respect to parameters a and b:
\[
F(\lambda, a, b) :=\left[\begin{array}{c}{\frac{a \cdot \lambda^{-5}}{\left(\frac{b}{\lambda}\right.} )} \\ {\frac{d}{d a} \frac{a \cdot \lambda^{-5}}{\frac{b}{\lambda}}} \\ {\frac{d}{d b} \frac{a \cdot \lambda^{-5}}{\left(e^{\frac{b}{\lambda}}-1\right)}}\end{array}\right]
\nonumber \]
Carry out nonlinear regression using Mathcad's genfit algorithm:
\[
\text{seed}:=\left(\begin{array}{c}{5 \cdot 10^{3}} \\ {10}\end{array}\right) \qquad P :=\text { genfit }(\lambda, \rho, \text { seed, } F) \qquad P=\left(\begin{array}{c}{4.715 \times 10^{3}} \\ {8.906}\end{array}\right) \qquad\left(\begin{array}{l}{a} \\ {b}\end{array}\right) :=P
\nonumber \]
Calculated radiation equation using output parameters:
\[
\rho_{\text { calc }}(\mathrm{L}, a, b)=\frac{a \cdot L^{-5}}{\left(e^{\frac{b}{L}}-1\right)}
\nonumber \]
Plot data and fit:
\[
\mathrm{L}=0.05, 0.1 \ldots 7
\nonumber \]
Calculate Planck's constant using the value of b, which is equal to (hc)/(kT).
\[
\mathrm{h} :=\frac{\mathrm{b} \cdot 10^{-6} \cdot 1.381 \cdot 10^{-23} \cdot 1595}{2.9979 \cdot 10^{8}} \qquad \mathrm{h}=6.544 \times 10^{-34}
\nonumber \]