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The spectrum of a blackbody has a peak wavelength of 2.80×10-7 meters. What is its temperature, in kelvins?

User Sonnia
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1 Answer

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Answer:
1.03* 10^4K

Step-by-step explanation:

Using Weins displacement law:


\lambda _(max)=(b)/(T)

where
\lambda _(max) = wavelength =
2.80* 10^(-7)m

b = constant =
2.897* 10^(-3)mK

T = Temperature in Kelvin = ?

Putting the values we get:


2.80* 10^(-7)m=(2.897* 10^(-3)mK)/(T)


T=1.03* 10^4K

Thus the temperature of the blackbody will be
1.03* 10^4K

User Piterio
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