Question

Express Planck's radiation law in terms of wavelength (a) as opposed to frequency. Hint: Start by performing a change of variable from v to 2 in equation 1.7 using c = va.

Answer 1

Answer:

$I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}$

Explanation:

The intensity of the radiation emitted by a black body with a certain temperature T and frequency $\nu$, is given by Planck's law:

$I(\nu,T)=\frac{2h\nu^3}{c^2}(\frac{1}{e^{\frac{h\nu}{kT}}-1})$

Considering the frequency range between $\nu$ and $\nu + \delta \nu$ and $dI$ the amount of energy emitted in this frequency range. Since an increase in frequency means a decrease in wavelength. Then:

$I(a,T)da=-I(\nu,T)d\nu\\I(a,T)=-\frac{d\nu}{da}I(\nu,T)$

Now recall that $\nu=\frac{c}{a}$, differentiate both sides:

$d\nu=-\frac{c}{a^2}da\\\frac{d\nu}{da}=-\frac{c}{a^2}$

Replacing this in previous equation:

$I(a,T)=\frac{c}{a^2}I(\nu,T)\\I(a,T)=\frac{c}{a^2}(\frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1})$

Rewriting $\nu^3$ as $\frac{c^3}{a^3}$ and $\nu$ as $\frac{c}{a}$

$I(a,T)=\frac{c}{a^2}(\frac{2hc^3}{a^3c^2}\frac{1}{e^{\frac{hc}{akT}}-1})\\I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}$

Finally, we obtain Planck's radiation law in terms of wavelength