Solving for temperature using Wiens Law

  • Thread starter FaraDazed
  • Start date
  • #1
FaraDazed
347
2

Homework Statement



A blackbody has a peak wavelength of 2x10^-6m , use the wavelength form of Wiens Law to calculate its temperature

Homework Equations


[tex]
\lambda_{peak}=\frac{hc}{5kT} \\
I_{\lambda}=\frac{2hc^2}{\lambda^5(e^{hc/ \lambda kT}-1)}
[/tex]


The Attempt at a Solution


I was not sure what equation to use but from the lecture booklet the two above are the only ones with temperature in them, the second one doesn't give me enough information as I don't know [itex]I_{\lambda}[/itex], however my answer just doesn't seem correct to me.

[tex]
\lambda_{peak}=\frac{hc}{5kT} \\
2×10^{-6}=\frac{(6.626×10^{-34})(2.997×10^8)}{5(1.38×10^{-23})T} \\
2×10^{-6}=\frac{1.986×10^{-25}}{(6.98×10^{-23})T} \\
T(6.98×10^{-23})(2×10^{-6})=1.986×10^{-25} \\
T(1.396×10^{-28})=1.986×10^{-25} \\
T=\frac{1.986×10^{-25}}{1.396×10^{-28}}=1422K
[/tex]
 
Last edited:

Answers and Replies

  • #2
Legaldose
74
6
There is an equation that you are missing that you need to use, it turns this into a 1 step problem. Wien's Law relates the peak wavelength to the objects temperature like so:

λpeak = [itex]\frac{.29cmK}{T}[/itex]

Is that equation not given?
 
  • #3
FaraDazed
347
2
There is an equation that you are missing that you need to use, it turns this into a 1 step problem. Wien's Law relates the peak wavelength to the objects temperature like so:

λpeak = [itex]\frac{.29cmK}{T}[/itex]

Is that equation not given?

Ah, I think so yeah but there is no K on the top though, it is given as

[tex]
\lambda_{peak}=\frac{2.898×10^{-3}}{T}
[/tex]

Using that I get

[tex]
T=\frac{2.898×10^{-3}}{2×10^{-6}}= 1449 Kelvin\\
[/tex]

So it is roughly the same, so the previous answer must have been correct :) Thanks.

EDIT: Wow I have made some silly input errors today :)
 
Last edited:
  • #4
Legaldose
74
6
Well the cmK was just supposed to be the units of the .29 :p
 
  • #5
FaraDazed
347
2
Well the cmK was just supposed to be the units of the .29 :p

Ah right, sorry :)
 

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