Planck's Law Problem: Solve for Vmax 3 Sig Digits

[nadeemo]

Homework Statement

Given Planck's Radiation Formula

Find the frequency (Vmax) at which energy density is at a maximum. This requires simple calculus and numerical solution of a simple transcendental equation.
You only need to find the answer to 3 significant digits.

Homework Equations

The Attempt at a Solution

So i derived the equation and set it equal to 0 and i got

V = [3kT / h][1 - 1/ (e^(hv/kT))]

But i can't seem to finish it because of the V in the exp...
also we know that
e^x = 1 + x + x^2/2! + x^3/3!+ ... and that somehow has to play a role in solving this...I think...

 

[Dick]

You have to do this numerically. Call hv/kT=X. Then your last equation is X/3=1-e^(-X). Write this as f(X)=1-X/3-e^(-X). You want to solve f(X)=0. Plot it. For values of X around 1 f(X) is positive, for values around 4 it's negative. So there must be a root in between. You could just solve it by bisection.

 

[nadeemo]

x/3 = 1 - e^(-x)

why is it 'x/3' ?, and not just x

 

[Dick]

x/3 = 1 - e^(-x)

why is it 'x/3' ?, and not just x

Because you have 3kT/h. Note the '3'.

 

[nadeemo]

wont it be 3x/v ??
since we have

hv / kt = x

andd in this part we have

3h/kt...wont that equal 3x / v

 

[Dick]

wont it be 3x/v ??
since we have

hv / kt = x

andd in this part we have

3h/kt...wont that equal 3x / v

Yes. And 3kt/h=3v/x. ?

 

[nadeemo]

hv / Kt = x
kt/hv = 1/x
3kt/hv = 3/x
3kt/h = 3v/ x

...i still don't understand where you get "x/3" from, please clarify

 

[Dick]

You had:

V = [3kT / h][1 - 1/ (e^(hv/kT))]

this is the same as:

Vh/(3kT)=[1 - 1/ (e^(hv/kT))]

Vh/(3kT)=x/3.