# Planck's law

1. May 30, 2009

### triac

1. The problem statement, all variables and given/known data
Hi! I'm trying to calculate the energy ratio of EM-radiation in the visible spectrum and EM-radiation in the entire spectrum, of a body with known temperature.

2. Relevant equations
Planck's law: http://feps.as.arizona.edu/outreach/images/Blackbody_gr_3.gif.[/URL]
However, I have a constant temperature and a variable wavelength, so I will let B(T) be I($$\lambda$$)

3. The attempt at a solution
I believe i can calculate the ratio (r) as: r=$$(\int_a^b\ I'(\lambda,T)d\lambda)/(\int_c^d\ I'(\lambda,T)d\lambda)$$ , where a and b mark the limits of the visible spectrum and c and d marks the limits of the entire spectrum. As I am unable to solve this algebraicly, I entered Planck's law on my calculator. However, the curve I go was completely wrong. Do you have any idea of what's wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Apr 24, 2017
2. May 30, 2009

### Hao

I believe you expect to see a curve similar to http://upload.wikimedia.org/wikipedia/commons/a/a2/Wiens_law.svg" [Broken]. The equation you have is correct, so I am not sure what could have gone wrong.

As for the integral you are looking for, the integral over energy density reduces to the standard form:

$$F = \int^{b}_{a} \frac{x^3}{e^x-1}dx$$

Unfortunately, this integral cannot be solved analytically, although when $$b \rightarrow \infty , a = 0$$ the value of $$F = \pi^4 / 15$$.

It will be necessary to use numerical integration to find the value of the integral for arbitrary limits. Fortunately, there are a number of such numerical solver online, as well as software packages if you have access to them. And of course, your graphical calculator.

If you were to list the steps you took and the output of your calculator, it will be possible to identify what went wrong.

Last edited by a moderator: May 4, 2017
3. May 31, 2009

### triac

Firstly, how do you mean it's reduced to the standard form F? How do I do the calculations using this F, is x the wavelength?

I enterd the function like this on my calculator: Y1=((2hc^2)/x^5)*(e$$^{hc/(xk*2700)}$$-1)$$^{-1}$$, where h is planck's constant, k is Boltzmann's constant, x is the wavelength, c is the lightspeed and 2700 is my temperature.

4. May 31, 2009

### Hao

When you find the integral
$$\int^{b_\lambda}_{a_\lambda}\frac{2 h c^2}{\lambda^5} \frac{1}{e^{\frac{h c}{\lambda k T}} - 1} d\lambda$$
a substitution $$x = \frac{h c}{\lambda k T}$$ will bring it to that standard form multiplied by a constant. The constant is pretty complex, and will include a factor $$T^4$$.

In this case, the limits a and b will be the x you get by putting in the boundary values of $$\lambda$$ into $$x = \frac{h c}{\lambda k T}$$.

The input is fine. If you didn't get the black body radiation curve, could you sketch the shape of the graph? One possibility is that the absolute values of x in such a curve are relatively small, and your graphical calculator might not be able to handle that. Additionally, your graphical calculator may be truncating the top half of the curve. Lastly, if your range to too wide, you will only see a plot of the decreasing exponential component as the peak only occurs relatively close to the origin.

I get the black body curve when I plot your expression in a graph plotting program.

5. May 31, 2009

### triac

You're right, it peaks around origo and drops exponentially. What graph plotting program are you using?

6. May 31, 2009

### Hao

I am using Mathematica. What you could do with your graphical calculator is is multiply your expression by a scaling factor like 0.0001 to bring the y coordinates into range.

7. May 31, 2009

### triac

D'oh, it still peaks around origo. I just wonder, can you calculate integrals in this mathematica? If yes, would you mind calculating the ratio between the integral from 400 nm to 750 nm and the integral from zero to infinity, with the temperature 2800 K? I would just like to see the answer to see if I'm on the right track.

8. May 31, 2009

### Hao

It is best to work in terms of the standard integral because it is dimensionless.

For $$\lambda \rightarrow 0, b \rightarrow \infty; \lambda \rightarrow \infty, a \rightarrow 0$$
$$F = \int^{\infty}_{0} \frac{x^3}{e^x-1}dx = \frac{\pi^4}{15}$$

For $$\lambda = 400nm, b = 12.87; \lambda = 750nm , a = 6.86$$
$$F = \int^{12.87}_{6.86} \frac{x^3}{e^x-1}dx = 0.528$$

The attached plot may be useful - it represents
$$F = \int^{x}_{0} \frac{s^3}{e^s-1}ds$$
F is the vertical axis, x is the horizontal axis.

#### Attached Files:

• ###### Nint.gif
File size:
5 KB
Views:
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Last edited: May 31, 2009
9. May 31, 2009

### triac

Thank you so much!

10. May 31, 2009

### triac

I guess I'll need some more help, if you have the time.
I found out my temperature was a bit lower, 2635 K, and that I was supposed to integrate from 380 nm to 750 nm, so I would need another calculation. However, it would be good if I understand that with the constants a and b and the standard integral. Do you think it would be posible for you to explain?

11. May 31, 2009

### Hao

Starting from
$$\int^{b_\lambda}_{a_\lambda}\frac{2 h c^2}{\lambda^5} \frac{1}{e^{\frac{h c}{\lambda k T}} - 1} d\lambda$$
, we let
$$\lambda = \frac{h c}{x k T}$$ or $$x = \frac{h c}{\lambda k T}$$

Substituting this in, we get for our limits (after relabeling a <-> b):

$${a_x} = \frac{h c}{b_\lambda k T}, {b_x} = \frac{h c}{a_\lambda k T}$$

we also get for the $$d\lambda$$
$$d\lambda = -\frac{h c}{x^2 k T} dx$$.

Putting all of this in, we get for the integral:

$$\int^{b_\lambda}_{a_\lambda}\frac{2 h c^2}{\lambda^5} \frac{1}{e^{\frac{h c}{\lambda k T}} - 1} d\lambda = \int^{a_x}_{b_x}\frac{2 h c^2 (x k T)^5}{(h c)^5}} \frac{1}{e^{x} - 1} (-\frac{h c}{x^2 k T}) dx = \int^{b_x}_{a_x}\frac{2 (k T)^4}{h^3 c^2} \frac{x^3}{e^{x} - 1} dx = \frac{2 (k T)^4}{h^3 c^2} \int^{b_x}_{a_x} \frac{x^3}{e^{x} - 1} dx$$

When my attachment in the previous post is up, you should be able to calculate answers for this integral for a fairly wide range of $$\lambda$$.

12. May 31, 2009

### triac

Thanks alot for your help, it's very appreciated!