Question

What temperature, in °C, is a blackbody whose emission spectrum peaks at (a)$300nm$ and (b)$3.00mm$?

Short answer

(a)$9393.67°C$(b)$693.67°C$

Step-by-step solution

(a) Given λ=300nm Now, We simply need to apply wien's displacement law :-

$\lambda \left(nm\right)=2.9x{10}^6(nm.k)\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.T$

$T=\raisebox{1ex}{$2.9x{10}^6(nm.K)$}\!\left/ \!\raisebox{-1ex}{$\lambda \left(nm\right)$}\right.$

width="196" height="28" role="math">T=2.9x106(nm.K)300 ($\lambda$=$300$)

$T=300T=9666.67K=9393.67°C$

(b) Given λ=3.00mm Again applying wein's displacement law :-

$$\begin{gathered} T=\raisebox{1ex}{$2.9x{10}^6(nm.K)$}\!\left/ \!\raisebox{-1ex}{$300$}\right. \\ T=966.67K \\ T=693.67°C \end{gathered}$$$(\lambda =3.00mm)$