Question

Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is 3 200 K.

Answer 1

Answer:

the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$

Explanation:

Given that:

the temperature = 3200 K

By applying  Wien's displacement law ,we have

$\lambda _m$T = 0.2898×10⁻² m.K

The peak wavelength of the emitted radiation at this temperature is given by

$\lambda _m$ = $\frac{0.2898*10^{-2} m.K}{3200 K}$

$\lambda _m$= $9.05625*10^{-7} \ m$

Hence, the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$