PHYS 275: Hw 3 Solutions
Chapter 3:
5-) Because is the power per unit area per unit wavelength, while
is the power per unit area per unit frequency. In other words, the argument
would have been valid if the left sides of these equations were the power per unit area, but they
are not. The two formulas are not equal to each other, rather they are related by .
8-) (a)
(b)
(c) # of photons per second per second.
11-) The radiation energy per volume per unit wavelength
Since we are asked to find the number of visible photons, let us write in terms of total number
of photons per volume per wavelength by dividing the by the energy of each photon
Thus . If we integrate the right-hand side over between 400-
750 nm for room temperature, we will get , the total number of visible photons per unit volume:
.
The total number in our room would be
photons, which is almost nonexistent!
