Answer to Question #227578 in Electric Circuits for Nicholas

Question

The length of a copper track on a printed circuit board has a cross sectional area of 5.0x10^-8. The current in the track is 3.5mA (3.5x10^-3). You are priveded with some useful information about copper:

1m^3 of copper has a massive of 8.9x10^3
54kg of copper contains 6.0x10^26
In copper, there is roughly one electron liberated from each copper atom.
a) Show that the electron number density "n" for copper is about 10^29m^-3

Answer 1

54kg atom contains number of atom =$6.0\times10^{26}$

1kg copper contain number of atom =$\frac{6.0\times10^{26}}{54}$

$8.9\times10^{3} kg$ atom contains number of atom=$\frac{6.0\times10^{26}}{54}\times(8.9\times10^3)$

$=9.88\times10^{28}=~10^{29}$ m^-3

electron number density "n" for copper is about (approximate ) $10^{29}m^{-3}$