Not quite.

Loosely described, the Richter assigns a numeric value to an earthquake based on the amplitude of vibrations recorded on a seismograph at a specific distance from the epicenter.

The formula is something like Magnitude = log (amplitude) + correction factor (1), where the correction factor accounts for the distance of the seismograph from the epicenter.

So, in an real situation, an amplitude of 23 mm (or 23000 microns) could result in a Richter magnitude of 5.
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html

To get a Richter magnitude of 9, keeping the correction factor the same as in the example, requires that A be on the order of 1,000,000 mm (1,000,000,000 µm), as you suggest (pretty much off the charts for the setup given).

The problem is that this is the amplitude as recorded on the seismograph paper *not the amplitude in terms of actual earth displacement.* To translate the recorded amplitude into the *real* amplitude is a bit more complex.

For a short answer, "The magnification of Richter's seismometer was about 2,800, so one millimeter on the paper record corresponds to about 0.36 microns of actual ground motion" (2). In this framework, a magnitude 9 would correspond to 1,000,000 mm, which translates into about 360 mm of actual ground motion.
Source(s):
http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html
http://tremor.nmt.edu/faq/how.html
http://www.scientificamerican.com/article.cfm?id=how-was-the-richter-scale

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