# Calculating Planck's integral for finite range of wavelength

• JulienB
In summary, the author calculates how much of the total radiated power of a light bulb at temperature 2300K is contained within 400nm and 750nm using Stefan-Boltmann law and Planck's law of radiation, respectively. He finds that the part of the power contained within the range of wavelength is estimated using the Planck's law of radiation and integrates from 400nm to 750nm. He finds that the result is 18.493 kW/m2.
JulienB

## Homework Statement

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##.

## Homework 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}##

## 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.

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.

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?

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.

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.

## 1. What is Planck's integral and why is it important in science?

Planck's integral is a mathematical formula used to describe the distribution of energy emitted by a blackbody at different wavelengths. It is important in science because it helps us understand the behavior of electromagnetic radiation and has many applications in fields such as astrophysics, thermodynamics, and quantum mechanics.

## 2. How do you calculate Planck's integral for a finite range of wavelength?

To calculate Planck's integral for a finite range of wavelength, you need to use the Planck's law equation and integrate it over the given range of wavelengths. This involves using calculus and numerical methods to solve the integral, which represents the total energy emitted by the blackbody in that specific range of wavelengths.

## 3. What are the units of Planck's integral?

The units of Planck's integral are energy per unit area per unit wavelength, typically expressed as joules per square meter per meter (J/m²m). It represents the amount of energy emitted by the blackbody at a specific wavelength per unit area.

## 4. How does Planck's integral change with temperature?

According to Planck's law, the total energy emitted by a blackbody increases with temperature and is inversely proportional to the wavelength. This means that as the temperature increases, the peak of the blackbody's emission spectrum shifts to shorter wavelengths and the overall energy emitted in a given wavelength range also increases.

## 5. What are some real-life applications of using Planck's integral?

Planck's integral has many practical applications, such as in the design of energy-efficient lighting systems, the study of thermal radiation in astrophysics, and the development of new materials for solar energy conversion. It is also used in medical imaging techniques, such as computed tomography (CT) and positron emission tomography (PET), to detect and measure radiation emitted by the human body.

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