= 2 
h c2 -5 / (exp(hc/kT) - 1).
F = 2 h c2 -5 / (exp(hc/kT) - 1).
or, combining the constants:
F = c1 -5 / exp(c2 /T) - 1),
where c1 = 2 hc2 = 3.7419 × 10-5 erg cm2 s-1 [ in cm]
and c2 = hc/k = 1.4288 cm °K.
In order to find the wavelength for which the emission is a maximum (at each temperature), we must take the derivative of this function and set it equal to 0, that is, to find the inflection point of the curve, the point at which a line tangent to the curve has a slope of 0. To do this, the first rule we will use is:
d(uv) = u dv + v du
where:
u = c1 -5 ;
du = -5 c1 -6 ;
v = [exp(c2 /T) - 1)]-1 and
dv = [c2/(2 T)] [exp(c2 /T)/(exp(c2 /T) - 1)2].
Combining these terms, we obtain:
0 = u dv + v du
= c1 -5 [c2/(2 T)] [exp(c2 /T)/(exp(c2 /T) - 1)] + [exp(c2 /T) - 1)](-5 c1 -6)]
Combining terms:
5c1/[ 6(exp(c2 /T) - 1)] =
c1c2exp(c2 /T)/[ 7 T (exp(c2 /T) - 1)]2
Which reduces to:
5 = c2 exp(c2 /T) / [ T (exp(c2 /T) - 1)] or
T =
[c2/5][exp(c2 /T) /(exp(c2 /T) - 1)]
We are almost there! c2/5 is easy to evaluate but the terms containing the exponentials are a bit more difficult because they still contain the wavelength variable. However, it is easy to see that the term containing the ratio of the 2 exponentials should be very nearly equal to 1 when we are near the peak of emission;
exp(c2 /T) 
exp(c2 /T) - 1
So let us iterate to obtain the final value for the right
hand side of the equation. As a first guess, let us take c2/5,
or 0.28777 as the value of T. Using this value to evaluate the exponentials,
we find a "correction factor" of 1.00681, yielding a new guess of
0.28973 for T.
With this value for T, the ratio of the exponentials is 1.00703.
Correcting the guess again, we obtain 0.28979. With this value the ratio
of the exponential terms is now 1.00705, yielding a corrected value of
0.28979. We have apparently converged in only 3 iterations! Therefore the final equation is:
T =
0.28979.
This equation is known as the Wien Displacement Law.