Question

Find the peak wavelength of the blackbody radiation emitted by the tungsten filament of a light bulb, which operated at 2000K

a) 1.45 μm
b) 1.65 μm
c) 2.32 μm
d) 2.75 μm

Answer 1

Using Wien's displacement law, the peak wavelength of the blackbody radiation from a light bulb filament operated at 2000 K is found to be approximately 1.45 µm, which corresponds to answer option (a).

Step-by-step explanation:

To find the peak wavelength of the blackbody radiation emitted by the tungsten filament of a light bulb operating at 2000 K, we can use Wien's displacement law.

The formula expressing the law is as follows:

λmax T = 2.898 × 10-3 m · K

To solve for λmax (the peak wavelength), we rearrange the formula to isolate λmax:

λmax = (2.898 × 10-3 m · K) / T

Plugging in the temperature of the tungsten filament (2000 K):

λmax = (2.898 × 10-3 m · K) / 2000 K

= 1.449 × 10-6 m

= 1.449 × 10-6 × 106 µm

= 1.449 µm

The peak wavelength is therefore closest to option (a) 1.45 µm.