Question #227578
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
1
Expert's answer
2021-08-22T15:48:22-0400

54kg atom contains number of atom =6.0×10266.0\times10^{26}

1kg copper contain number of atom =6.0×102654\frac{6.0\times10^{26}}{54}

8.9×103kg8.9\times10^{3} kg atom contains number of atom=6.0×102654×(8.9×103)\frac{6.0\times10^{26}}{54}\times(8.9\times10^3)

=9.88×1028= 1029=9.88\times10^{28}=~10^{29} m-3

electron number density "n" for copper is about (approximate ) 1029m310^{29}m^{-3}


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