Answer:
[tex]I(a,T)=\frac{2hc^2}{a^5}\frac{1}{e^{\frac{hc}{akT}}-1}[/tex]
Explanation:
The intensity of the radiation emitted by a black body with a certain temperature T and frequency [tex]\nu[/tex], is given by Planck's law:
[tex]I(\nu,T)=\frac{2h\nu^3}{c^2}(\frac{1}{e^{\frac{h\nu}{kT}}-1})[/tex]
Considering the frequency range between [tex]\nu[/tex] and [tex]\nu + \delta \nu[/tex] and [tex]dI[/tex] the amount of energy emitted in this frequency range. Since an increase in frequency means a decrease in wavelength. Then:
[tex]I(a,T)da=-I(\nu,T)d\nu\\I(a,T)=-\frac{d\nu}{da}I(\nu,T)[/tex]
Now recall that [tex]\nu=\frac{c}{a}[/tex], differentiate both sides:
[tex]d\nu=-\frac{c}{a^2}da\\\frac{d\nu}{da}=-\frac{c}{a^2}[/tex]
Replacing this in previous equation:
[tex]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})[/tex]
Rewriting [tex]\nu^3[/tex] as [tex]\frac{c^3}{a^3}[/tex] and [tex]\nu[/tex] as [tex]\frac{c}{a}[/tex]
[tex]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}[/tex]
Finally, we obtain Planck's radiation law in terms of wavelength
