I am currently studying Introductory Semiconductor Device Physics by Parker. Chapter 2.5 The concept of effective mass gives the following example:

For GaAs, calculate the typical (band-gap) photon energy and momentum, and compare this with a typical phonon energy and momentum that might be expected with this material.

The band gap of GaAs is about $1.43 \ \text{eV}$ (so take this for the photon energy). Use equation (2.2) to estimate a typical photon wavelength:

Wavelength (microns) $= 1.24/1.43 = 0.88 \ \text{$\mu$m}$

The photon momentum can be calculated from equation (2.6):

Momentum $= h/0.88 \times 10^{-6} = 7.53 \times 10^{-28} \ \text{kg m s$^{-1}$}$ Now let's do the same sort of calculation for the phonon. We can use the same basic equations but we need to change a couple of values. Instead of the velocity of light, the phonons will travel at the velocity of sound in the material, approximately $5 \times 10^3 \ \text{m s$^{-1}$}$. Also, instead of the (long) photon wavelength the phonon wavelength is of the order of the material's lattice constant, which is $5.65 \times 10^{-10} \ \text{m}$ in this case. From the equations you should see immediately that the phonon energy is going to be pretty small, because the velocity has dropped from the velocity of light to the velocity of sound, but the phonon momentum is going to be pretty big, because we are dividing by the lattice parameter, which is very small compared to the photon wavelength. Putting this altogether gives,

Phonon energy $= 0.037 \ \text{eV}$

Phonon momentum $= 1.17 \times 10^{-24} \ \text{kg m s$^{-1}$}$

Equations (2.2) and (2.6) are as follows:

$$\text{ENERGY (in eV)} = 1.24/\text{WAVELENGTH (in microns)} \tag{2.2}$$

$$p = \dfrac{h}{\lambda} \tag{2.6}$$

We are also given the energy-momentum equation:

$$E = \dfrac{\hbar^2 k^2}{2m},$$

where $\hbar$ is the reduced Planck constant and $k$ is the propagation constant of a wave.

I was able to calculate everything except for the phonon energy. How did the author calculate the phonon energy? Did they somehow use the formula $\text{momentum} = \text{mass} \times \text{velocity}$?

I would greatly appreciate it if people would please take the time to clarify this.


1 Answer 1


The author takes the phonon wavelength to be $\lambda=5.65\times10^{-10}\ \text{m}$. By "momentum", the author is referring to the crystal momentum of the phonon, which is $\hbar k=\frac{h}{\lambda}=1.17\times10^{-24}\ \text{kg}\text{m}^{-1}\text{s}^{-1}$. At long wavelengths, the energy of acoustic (low energy) phonons is approximately $vp$, where $p$ here is the crystal momentum and $v$ is the speed of sound. With $v=5\times10^3\ \text{ms}^{-1}$, you get $E=0.037\ \text{eV}$.

Note however that the speed of sound in most materials depends on the direction and phonon polarization. Also, the relation $E=vp$ is not valid for optical phonons or acoustic phonons with short wavelength.


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