Question

The average energy of Planck's oscillator is given by:

• A. \(kT\)
• B. \(\dfrac{3}{2}kT\)
• C. \(\dfrac{h\nu}{e^{h\nu/kT} - 1}\)
• D. \(\dfrac{kT}{2}\)

Hint:

Average over discrete energies \(nh\nu\) with Boltzmann weights; sum the geometric series to get \(h\nu/(e^{h\nu/kT}-1)\).

Answer 1

The Correct Option is C

Step 1: A Planck oscillator can only have discrete energies \(E_n = nh\nu\), with \(n = 0, 1, 2, \dots\). The average energy is
\[\langle E \rangle = \frac{\sum_{n=0}^{\infty} n h\nu\, e^{-nh\nu/kT}}{\sum_{n=0}^{\infty} e^{-nh\nu/kT}}.\]
Step 2: Let \(x = e^{-h\nu/kT}\). The denominator is a geometric series \(\sum x^n = \dfrac{1}{1-x}\).

Step 3: The numerator is \(h\nu \sum n x^n = h\nu \dfrac{x}{(1-x)^2}\).

Step 4: Dividing:
\[\langle E \rangle = h\nu \frac{x}{1-x} = \frac{h\nu}{\tfrac{1}{x} - 1} = \frac{h\nu}{e^{h\nu/kT} - 1}.\]
\[\boxed{\langle E \rangle = \frac{h\nu}{e^{h\nu/kT} - 1}}\]