Ay 20 – Set 4: Black Body Radiation
Problem 2a
Primary author: Joanna Robaszewski
Secondary author: Cassi Lochhaas
Abstract
This problem demonstrates the relationship between the black body intensity
and the black body intensity
by examining the units of each.
Introduction
A black body is something that emits radiation perfectly so that the properties of the emitted light are solely dependent on the temperature of the body. The intensity of a black body is a measurement of energy per time, per frequency, per area. We can use either the frequency ν or the wavelength λ to look at the intensity at some particular frequency. So we get the equations:
Where h is 6.6 x 10-27 erg*s and k is 1.4 x 10-16 ergs K-1 , T is the temperature of the black body, and c is the speed of light.
Questions and Results
to
Let’s start by asking what are the units on B_nu?
Which simplifies to:
So to go from the units of B_nu to the units of B_lambda, what do we have to multiply B_nu by?
We need to get another unit of centimeters in the denominator, along with seconds squared.
We know that frequency has units of s-1 so we should try multiplying by the frequency squared. If we do this we get the following units:
This is slightly closer to what we are looking for, but we still need centimeters cubed to end up in the denominator.
Let’s consider the relationship:
We want λ in terms of ν so if we solve for λ we get:
which has units of cm. We want another factor of cm in the denominator, so if we take the reciprocal, we will get:
with units of cm-1.
So if we multiply B_nu by:
which has units of:
we get:
which are the units of B_lambda!
And all we had to do was multiply
Here is a picture of a physicist's representation of a cow in space as a spherical black body. Please note that the jet packs do not emit any radiation in this particular situation and there are no stars in the background because the exposure time was not very long.
I love the cow!
ReplyDeleteAs for your solution, you should think about the clue I gave you on the worksheet. The energy in an interval d\nu must be the same as the energy in an interval d\lambda, all other factors (time, area, solid angle) being equal. What do you get when you take this fact into account?
How we got to c = lambda*nu was by considering that the energy in an interval d_nu must be the same as the energy in an interval d_lambda. We thought that if this was to be true then since E=hc/lambda and E = h*nu, dE = hc/d_lambda and dE = h*d_nu. These dEs must be equal so hc/d_lambda = h*d_nu. Simplifying gives c = d_lambda * d_nu.
ReplyDeleteSince the energy in a wavelength interval is equal to the energy in the corresponding frequency interval, you can make the statement:
ReplyDeleteB_lambda * d_lambda = B_nu * d_nu
So B_lambda = B_nu * (d_nu/d_lambda).
What solution do you get by applying this?
ps - super cute cow!
ReplyDelete