Question

How many conduction electrons are there in a 4.00 mm diameter gold wire that is 30.0 cm long? and How far must the sea of electrons in the wire move to deliver -21.0 nC of charge to an electrode?

Answer 1

volume = ( pi d^2 / 4) (L)

= (pi (2 x 10^-3)^2) (0.30)

= 3.77 x 10^-6 m^3

mass = density x volume = (19320)(3.77 x 10^-6)

m = 0.0728 kg

moles = m / M = (72.83 grams) / (200)

= 0.365 mol

number of atoms = 0.365 x6.022 x 10^23 = 2.193 x 10^23 atoms

there is 1 conduction electron per atom.

total conduction electron = 2.193 x 10^23 electrons ......Ans

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number of electrson in the wire,

N = (21 x 10^-9)/(1.6 x 10^-19)

N = 13.125 x 10^10

so r = N L / n

r = (13.25 x 10^10) (0.30) / (2.193 x 10^22)

r = 1.80 x 10^-12 m Or 1.80 pm