Chapter 3: Problem 13
Equation \((3-1)\) expresses Planck's spectral energy density as an energy per range \(d f\) of frequencies. Quite of ten, it is more convenient to express it as an energy per range \(d \lambda\) of wavelengths, By differentiating \(f=c / \lambda,\) we find that df \(=-c / \lambda^{2} d \lambda\). Ignoring the minus sign (we are interested only in relating the magnitudes of the ranges \(d f\) and \(d \lambda\) ). show that, in terms of wavelength. Planck's formula is \(\frac{d U}{d \lambda}=\frac{8 \pi V h c}{e^{h c / \lambda k_{B} T}-1} \frac{1}{\lambda^{5}}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.