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Quantum Mechanics (Wien’s displacement law)

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Homework Statement


Show that the maximum of the Planck energy density occurs for a wavelength of the form λmax = b/T, where T is the temperature and b is a constant that needs to be estimated.


Homework Equations


Planck energy density
u (v,T) = 8πv2 / c3 * hv / ehv/kT-1

The Attempt at a Solution



v= c / λ
dv = |dv / (dλ)| dλ = (c/λ2) dλ

I get up to this bit and I'm stuck...

v= c / λ
c = v*λ
∴ v = v*λ / λ
dv = dv*dλ / dλ
dv = (dv / dλ) * dλ

I'm confused how dv = |dv / (dλ)| dλ turns into (c/λ2) dλ. Sorry if my question is a bit vague -- haven't posted on here in a while and I just started taking QM.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,394
1,802

Homework Statement


Show that the maximum of the Planck energy density occurs for a wavelength of the form λmax = b/T, where T is the temperature and b is a constant that needs to be estimated.


Homework Equations


Planck energy density
u (v,T) = 8πv2 / c3 * hv / ehv/kT-1

The Attempt at a Solution



v= c / λ
dv = |dv / (dλ)| dλ = (c/λ2) dλ

I get up to this bit and I'm stuck...

v= c / λ
c = v*λ
∴ v = v*λ / λ
dv = dv*dλ / dλ
dv = (dv / dλ) * dλ


I'm confused how dv = |dv / (dλ)| dλ turns into (c/λ2) dλ. Sorry if my question is a bit vague -- haven't posted on here in a while and I just started taking QM.
The absolute value marks are wrong in your equation.
d means differential, and dv/dλ is the derivative of v with respect to λ. d is not a multiplicative factor!
v is function of λ, and u(v,T)=F(v(λ),T). v=c/λ(v). You have to apply the chain rule to find the λ, where the u( λ) plot has its maximum:
dF/dλ = df/dv dv/dλ.
v=c/ λ, what is its derivative with respect to λ?
 
  • #3
594
12
v=c/ λ, what is its derivative with respect to λ?

dv = ( c / λ2 )
 
  • #4
594
12
v=c/ λ, what is its derivative with respect to λ?
dv = ( c / λ2 ) dλ
 
  • #5
ehild
Homework Helper
15,394
1,802
v=c/ λ, what is its derivative with respect to λ?

dv = ( c / λ2 )
No, the derivative is written as ##\frac{dv}{dλ}##
In the derivative, you miss a minus sign.
 

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