gnome
Mar11-04, 06:59 PM
Basically this problem is to derive Wien's displacement law from Planck's law.
Specifically:
a) Show that there is a general relationship between temperature and λmax stating that Tλmax = constant
and
b) Obtain a numerical value for this constant
[Hint: Start with Planck's radiation law and note that the slope of u(λ,T) is zero when λ = λmax ]
So I obediently try to differentiate u(λ, T) with respect to λ:
I'll let C1 = 8Πhc and C2 = hc/(kBT)
u(\lambda, T) = {C_1}\lambda^{-5} \left( e^\frac{C_2}{\lambda} - 1 \right)^{-1}
\frac{\partial u}{\partial \lambda} =
C_1 \left((\lambda^{-5})(-1)({e^{\frac{C_2}{\lambda}} - 1)^{-2}({e^\frac{C_2}{\lambda}})(-\frac{C_2}{\lambda^2}) + (e^{\frac{C_2}{\lambda}} - 1)^{-1}(-5)\lambda^{-6}\right) = 0
I divide through by C1 & get
\frac{C_2e^\frac{C_2}{\lambda}}{\lambda^7({e^\frac {C_2}{\lambda}} -1)^2} - \frac{5}{\lambda^6({e^\frac{C_2}{\lambda}} -1)} = 0
\frac{C_2{e^\frac{C_2}{\lambda}} - 5\lambda({e^\frac{C_2}{\lambda}} - 1)}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} = 0
C_2{e^\frac{C_2}{\lambda}} - 5\lambda{e^\frac{C_2}{\lambda}} + 5\lambda = 0
let x = \frac{C_2}{\lambda} = \frac{hc}{{k_B}T\lambda}
x{\lambda}e^x - 5{\lambda}e^x + 5{\lambda} = 0
x = 5 - \frac{5}{e^x} = 5(1 - e^{-x})
Now I substitute back for x = hc/(kTλ) and get:
\frac{hc}{{k_B}T{\lambda}} = 5(1 - {e^\frac{-hc}{{k_B}T{\lambda}}})
Oy!
This, amazingly, is exactly the expression I'm supposed to end up with (for part a), according to the answer in the book.
But how does this show that
T\lambda = constant
and how do I solve for this constant?
Specifically:
a) Show that there is a general relationship between temperature and λmax stating that Tλmax = constant
and
b) Obtain a numerical value for this constant
[Hint: Start with Planck's radiation law and note that the slope of u(λ,T) is zero when λ = λmax ]
So I obediently try to differentiate u(λ, T) with respect to λ:
I'll let C1 = 8Πhc and C2 = hc/(kBT)
u(\lambda, T) = {C_1}\lambda^{-5} \left( e^\frac{C_2}{\lambda} - 1 \right)^{-1}
\frac{\partial u}{\partial \lambda} =
C_1 \left((\lambda^{-5})(-1)({e^{\frac{C_2}{\lambda}} - 1)^{-2}({e^\frac{C_2}{\lambda}})(-\frac{C_2}{\lambda^2}) + (e^{\frac{C_2}{\lambda}} - 1)^{-1}(-5)\lambda^{-6}\right) = 0
I divide through by C1 & get
\frac{C_2e^\frac{C_2}{\lambda}}{\lambda^7({e^\frac {C_2}{\lambda}} -1)^2} - \frac{5}{\lambda^6({e^\frac{C_2}{\lambda}} -1)} = 0
\frac{C_2{e^\frac{C_2}{\lambda}} - 5\lambda({e^\frac{C_2}{\lambda}} - 1)}{\lambda^7({e^\frac{C_2}{\lambda}} -1)^2} = 0
C_2{e^\frac{C_2}{\lambda}} - 5\lambda{e^\frac{C_2}{\lambda}} + 5\lambda = 0
let x = \frac{C_2}{\lambda} = \frac{hc}{{k_B}T\lambda}
x{\lambda}e^x - 5{\lambda}e^x + 5{\lambda} = 0
x = 5 - \frac{5}{e^x} = 5(1 - e^{-x})
Now I substitute back for x = hc/(kTλ) and get:
\frac{hc}{{k_B}T{\lambda}} = 5(1 - {e^\frac{-hc}{{k_B}T{\lambda}}})
Oy!
This, amazingly, is exactly the expression I'm supposed to end up with (for part a), according to the answer in the book.
But how does this show that
T\lambda = constant
and how do I solve for this constant?