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

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

Step-by-step explanation:

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

`I(\\u,T)=(2h\\u^3)/(c^2)(\frac{1}{e^{(h\\u)/(kT)}-1})`

Considering the frequency range between
`\\u` and
`\\u + \delta \\u` 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(\\u,T)d\\u\\I(a,T)=-(d\\u)/(da)I(\\u,T)`

Now recall that
`\\u=(c)/(a)`, differentiate both sides:

`d\\u=-(c)/(a^2)da\\(d\\u)/(da)=-(c)/(a^2)`

Replacing this in previous equation:

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

Rewriting
`\\u^3` as
`(c^3)/(a^3)` and
`\\u` as
`(c)/(a)`

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

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