The tungsten filament of an incandescent light bulb has a temperature of approximately 3000 K. The emissivity of tungsten is approximately 1/3, and you may assume that it is independent of wavelength. At what value of the photon wavelength does the peak in the bulb’s spectrum occur?

I am a little confused with this question. My thought is that the answer should just $\lambda = b/T$, which is the Wien's displacement law. BUT if we substitute the values, we got the wrong answer (given by the book, the answer is: ($1.7 \space \mu m$)

I have thought that, maybe, the emissivity not being $e=1$ modifies the wien's law, but i can't see why:

Since $u(\lambda) = \frac{ 8\pi v^2 (hv) ^3}{c(hc)^3 (e^{hv/kt}-1)} = $, my guess is that for non perfectly emitters as in this case, $u' = u1/3$, but to get the peak we just evaluate $du/d\lambda = 0$, and all constants can be canceled, so why is the answer different than $\lambda = b/T$?


1 Answer 1


It's a badly formulated question. You need to know what "spectrum" means.

If it means the flux per unit frequency, then the peak of the spectrum is at $$ \nu_{\rm max} = 2.82 \frac{k_B T}{h}\ . $$

If you work out the wavelength corresponding to this, then it is $1.7 \mu$m.

If the "spectrum" is defined as flux per unit wavelength, then you get a different answer because (see for example Peaks different in wavelength and frequency space and the many duplicates) $$ f(\lambda) = f(\nu)\frac{d\nu}{d\lambda}\ . $$


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