# Spectral radiance of the sun for 400nm

1. Dec 26, 2013

### bvbellomo

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

Using Planck's law, calculate the spectral radiance of 400nm sunlight arriving at earth. Assume 1/3600 steradians of view and 5800 K of temperature.

2. Relevant equations

Planck's law

3. The attempt at a solution

I am not the best at typing formulas.

Lets start what we are raising e to the power of. We have the Planck constant times the speed of light over the wavelength times the Boltzmann constant times temperature.

Planck constant = 6.63E-34
speed of light = 3E8
wavelength = 400nm = 400e-9
Boltzmann constant = 1.3806488E-23
6.63E-34 * 3E8 / (400e-9 * 1.3806488E-23 * 5800) = 6.2095993290031215

Raising e to that power gives me 497.50187677476521, subtracting 1 gives me 496.50187677476521

So I have 2 * h * c2 / y5 divided by that value
Where h is my Planck constant as above
c2 is the speed of light squared, as above
y5 is my wavelength to the fifth, as above

Which gives 2 * 6.63E-34 * 3E8 * 3E8 / (400e-9 * 400e-9 * 400e-9 * 400e-9 * 400e-9)
which is 11654296874999994

Divide by what I had above gives

23472815351083 watts / square meter / steradian

or 6520226486 watts / square meter

or 6.5 gigawatts / square meter. As I am not instantly vaporized as I write my answer, I know I made a mistake. But where?

2. Dec 26, 2013

### Staff: Mentor

Check the units, your calculation does not give W/m2.
Another way to see this: W/m2 is not a spectral radiance (you calculated something for 400nm, so the result has to have some relation to 400nm...).

3. Dec 26, 2013

### bvbellomo

Thanks, but I am still confused. Am I correct that I have 23472815351083 watts / square meter / steradian?

Googling a graph of the sun's output, I can see the correct answer for the question as I worded it is close to 1.6 watts / square meter of 400nm light. Which puts me off by a factor of 4e9, which is remarkably similar to the 400e9 wavelegth. Am I heading in the right direction, or is this a coincidence?

I understand that I need to integrate this from 400nm to 700nm if I want to know actual watts / square meter of 400nm to 700nm light over that range. However the integral of an incorrect function is usually an incorrect answer, so I want to leave out the calculus until I have the algebra working.

4. Dec 26, 2013

### Staff: Mentor

No.
To see that this is wrong, do the same calculation with 100nm instead of 400nm. What do you get?
You cannot get two different values for the total radiation if you calculate it in two different ways...

That is not right. If you look at "exactly" 400nm, the power is zero. You always need a wavelength range to get a non-zero power. Check the range that is given there, I guess the value of 1.6W/m2 is per nanometer of wavelength.
The 4 could be a coincidence, the e9 is a hint what you did wrong.

Integrate from 400nm to 401nm, that should give a reasonable result as soon as you fix the error in the units.

5. Dec 26, 2013

### bvbellomo

I think I figured this out. Intuitively, I understand the function I am integrating is really per nanometer of wavelength, but using SI units, my function is actually watts / meter (of wavelength) / square meter (of area) / steradian.

6. Dec 26, 2013

Exactly.