# Calculating Planck's integral for finite range of wavelength

1. Apr 22, 2017

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I am asked to calculate how much of the total radiated power of a light bulb at temperature $T=2300$K is contained within $400$nm and $750$nm. I am also given the average emissivity of tungsten $\epsilon_\text{ave}=0.288$ and the emissivity within the range of wavelength $\epsilon_\text{visible} \approx 0.45$.

2. Relevant equations

Stefan-Boltmann law: $R=\epsilon(\lambda) \sigma T^4$
Planck's law of radiation: $\frac{dR}{d\lambda} = \frac{2\pi h c^2}{\lambda^5} \frac{\epsilon(\lambda)}{\exp (\frac{hc}{\lambda k T}) -1}$

3. The attempt at a solution

First I calculated the total power per unit area $R_\text{ave}$ using Stefan-Boltzmann law:

$R_\text{ave} = \epsilon_\text{ave} \sigma T^4 = 457$kW$\cdot$m$^{-2}$

I spent some time trying to think of how to estimate the part of the power contained within the range of wavelength. The only solution I found so far is to use the Planck's law of radiation and integrate from $\lambda_1 = 400$nm to $\lambda_2 = 750$nm, but that isn't easy:

$dR_\text{visible} = \frac{2\pi h c^2}{\lambda^5} \frac{\epsilon_\text{visible}}{\exp (\frac{hc}{\lambda k T}) -1} d\lambda$
$\implies R_\text{visible} = \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \int_{u_1}^{u_2} \frac{u^3}{\exp(u)-1} du$

Here I have used substitution with $u=\frac{hc}{\lambda kT}$ like one would normally do with the integral from $0$ to $\infty$. Then with $\frac{1}{\exp(u)-1} = \sum_{n=1}^{\infty} e^{-nu}$:

$R_\text{visible} = \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \sum_{n=1}^{\infty} \int_{u_1}^{u_2} u^3 e^{-nu} du$
$= \frac{2 \pi (kT)^4}{h^3 c^2} \epsilon_\text{visible} \sum_{n=1}^{\infty} \bigg( (\frac{u_1^3}{n} + \frac{u_1^2}{n^2} + \frac{6 u_1}{n^3} +\frac{6}{n^4}) e^{-n u_1} - (\frac{u_2^3}{n} + \frac{u_2^2}{n^2} + \frac{6 u_2}{n^3} + \frac{6}{n^4}) e^{-n u_2} \bigg)$

after performing integration by parts several times and inserting the limits of integration (pfiuu). I calculated the first 5 terms of the sum using Matlab, and this gives me the result

$R_\text{visible} = 0.011$ W$\cdot$m$^{-2}$

which I find way too low to be true. Did I make a mistake somewhere? Some websites seem to confirm my calculation of the integral (for example here: http://www.spectralcalc.com/blackbody/inband_radiance.html), but maybe there is another way to calculate/approximate this calculation? I think I cannot use Rayleigh-Jeans approximation, since the peak of radiation is at intensity $\lambda_\text{max} = 1.26 \mu$m, which means that the Rayleigh-Jeans approximation goes towards $\infty$ on the left of the peak, which is not the case with Planck.

Any advice here? I think I might have gone too far with the integral, when maybe there is another path to the solution.

Thank you very much in advance.

Julien.

2. Apr 22, 2017

### JulienB

Actually I made a typing mistake in Matlab. The result is now $R_\text{visible}= -18.493$kW$\cdot$m$^{-2}$ which is better than before, except for the minus sign. Maybe I inverted my limits of integration by mistake somewhere.

3. Apr 22, 2017

### JulienB

I found the mistake, it was in my first substitution: either I forgot a $-$ or to invert the limits of integration. The result is now $R_\text{visible} = 18.493$kW$\cdot$m$^{-2}$ which represents about $4{\%}$ of the total electromagnetic power. I think it is a reasonable figure, what do you guys think?

Still I would be curious to know if there is another method of estimating this value. Maybe something with Wien displacement law?

Julien.

4. Apr 22, 2017

### vela

Staff Emeritus
Since you're using Matlab, you could just have it evaluate the integral numerically. I got 21.9 kW/m2.

You could try estimating the integral in different ways, but I think you always have to integrate $dR/d\lambda$ in some way.

5. Apr 23, 2017

### JulienB

@vela Thanks for your answer. Indeed I didn't know how to numerically integrate using Matlab until now. Thanks a lot, I got now the same result as you!

Julien.