## Re: converst from 4.0 x 10e21 cm^-3 to mole

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Posted by Roger Brennan (66.122.81.186) on March 30, 2010 at 13:40:26:

In Reply to: Re: converst from 4.0 x 10e21 cm^-3 to mole posted by napatporn on March 29, 2010 at 04:05:14:

: : : How can I coverst carrier concentration from 4.0 x10e21 cm^-3 to mole
: : Thank you

Hi,

Thank you for the interesting question.

I have been working in semiconductors since 1965 and this is the first time this has come up.

A mole is the atomic weight or molecular weight in grams. It is equal to 6.022 x 10^23 entities (Avagodro's number).

Check out Wikipedia for a short discussion of Avagodro's number. Here's part of it:
"
In chemistry and physics, the Avogadro constant (symbols: L, NA) is the number of "elementary entities" (usually atoms or molecules) in one mole, that is (from the definition of the mole), the number of atoms in exactly 12 grams of carbon-12. "

With this number, we can calculate how many cubic centimeters it takes to make a mole with the carrier concentration is 4E21cm-3

Na = 6.022E23 / mole

1 mole = 6.022E23

converting to cm-3 = 6.022E23 * cm-3 /4E21 = ~ 1.505 E2 cm-3 or about 150cm-3 of silicon (~0.15 liters)!

We usually reserve the term liters for liquids and gasses but I see no harm down using it for the volume of a solid.

Then you have 1/0.15 = 6.7 M solution. (A good working definition of 1 molar solution (1M) is one mole per 1 liter of solution.)

One note of caution: 4E21 cm-3 is a VERY high carrier concentration. In our experience, practical limits are more like 1 or 2E20 cm-3.

Regards, Roger

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