# 1.88: Planck's Radiation Equation Fit to Experimental Data - Another Algorithm


$\mathrm{n} :=42 \qquad \mathrm{i} :=1 . . \mathrm{n} \nonumber$

 $$\rho_{i} :=$$ $$\lambda_{i} :=$$ 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).

$\mathrm{h} :=\frac{\mathrm{b} \cdot 10^{-6} \cdot 1.381 \cdot 10^{-23} \cdot 1595}{2.9979 \cdot 10^{8}} \qquad \mathrm{h}=6.544 \times 10^{-34} \nonumber$

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