00:01
Okay.
00:03
In this problem, we have a system of different temperatures in parts a, b, and c.
00:08
And we were asked to find the wavelength of peak spectral emission given a plonk distribution.
00:17
So what we're asked to find is a lambda m for max.
00:29
So the main equation we'll need to use is the wind displacement law, which states simply that lambda m.
00:46
Is equal to hc over the empirical 4 .965 k times t where h is a plane constant c is speed of light and k is the boltzman constant so this is all we really need to use so let's just jump into it.
01:13
So part a we have t is equal to only three kelvin plugging that in lambda m is hc two constants all over 4 .965 times 3 kelvin.
01:39
Plug in the h constant and the c constant.
01:43
We end up with a result of 9 .66 times 10 to the negative 4 meters.
01:54
So this has a very, very long wavelength.
01:56
This is definitely way in the infrared...