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}$}$

Equation (2.2) is

$$\text{Energy (in $eV$)} = 1.24/\text{Wavelength (in microns)},$$

which relates the energy of a photon to its wavelength in microns ($10^{-6} m$ or $\mu m$). So we get

$$\dfrac{1.24}{1.43} = 0.87 \mu m$$

Equation (2.6) is

$$p = \dfrac{h}{\lambda},$$

where $h = 6.63 \times 10^{-34}$ is Planck's constant, and it relates momentum to wavelength. So we get

$$p = \dfrac{6.63 \times 10^{-34}}{0.87 \times 10^{-6}} = 7.62 \times 10^{-28} \ \ \text{(Apparently the units of the denominator is seconds?)}$$

As you can see, it seems that the author's calculations might be slightly inaccurate, but that's not what I'm concerned about.

My research indicates that the units of Planck's constant are joule-seconds, which, according to this Wikipedia article, is $kg m^2 s^{-1}$ in SI base units. So how did the author's calculations for the band gap photon momentum for GaAs result in $= h/0.88 \times 10^{-6} = 7.53 \times 10^{-28} \ kg m s^{-1}$? Where are we dividing by a meter to get $m$ instead of $m^2$ in the result?

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

  • 1
    $\begingroup$ The wavelength is $0.88\times 10^{-6}\ {\rm m}$ and you are dividing by the wavelength. $\endgroup$
    – The Photon
    Dec 30, 2019 at 17:17

1 Answer 1


Where are we dividing by a meter to get m instead of m2 in the result?

A wavelength is the distance between successive peaks of the field, and has units of length.

  • $\begingroup$ You're right; evidently (see my comment in the calculation for momentum), I somehow confused myself and thought we had microseconds instead of micrometers. Thanks for that. $\endgroup$ Dec 30, 2019 at 17:22

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