Question

What temperature, in °C, is a blackbody whose emission spectrum peaks at (a) $300nm$and (b) localid="1649698615260" $3.00\mu m$?

Short answer

Required temperatures are:

(a) $T=9393.67°C$

(b) $T=693.67°C$

Step-by-step solution

Part (a): Step 1: Given information

We have been given the wavelength as:

$\lambda =300nm$

Part (a): Step 2: Calculation

Using Wien's law,

$\lambda \left(nm\right)=\frac{2.9·{10}^6(nm·K)}{T}$

Here $\lambda$is wavelength in nanometers and $T$is temperature in Kelvin.

We get the expression for T: $T=\frac{2.9·{10}^6(nm·K)}{\lambda \left(nm\right)}$

Substituting $\lambda =300nm$,

$$\begin{gathered} T=\frac{2.9·{10}^6(nm·K)}{300} \\ T=9666.67K \\ T=9393.67°C \end{gathered}$$

Part (b): Step 1: Given Information

We have been given the wavelength as:

$\lambda =3.00\mu m$

Part (b): Step 2: Calculation

Using Wein's Law:

$$\begin{gathered} T=\frac{2.9·{10}^6(nm·K)}{3000} \\ T=966.67K \\ T=693.67°C \end{gathered}$$