Solving for temperature using Wiens Law

[FaraDazed]

Homework Statement

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

Homework Equations

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

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 doesnt give me enough information as I dont know $I_{\lambda}$, however my answer just doesnt seem correct to me.

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

 

[Legaldose]

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:

λ_peak = $\frac{.29cmK}{T}$

Is that equation not given?

 

[FaraDazed]

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:

λ_peak = $\frac{.29cmK}{T}$

Is that equation not given?

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

$$\lambda_{peak}=\frac{2.898×10^{-3}}{T}$$

Using that I get

$$T=\frac{2.898×10^{-3}}{2×10^{-6}}= 1449 Kelvin\\$$

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 :)

 

[Legaldose]

Well the cmK was just supposed to be the units of the .29 :p

 

[FaraDazed]

Well the cmK was just supposed to be the units of the .29 :p

Ah right, sorry :)