A 90mmdiameter copper ball is charged to 90 nC What fraction

A 9.0-mm-diameter copper ball is charged to 90 nC. What fraction of its electrons have been removed? The density of copper is 8900 kg/m^3.

Solution

volume of this ball = 4 pi r^3 /3

   = 4 x pi x (0.009/2)^3 / 3 = 3.817 x 10^-7 m^3

mass of ball = density x volume

= 8900 x 3.817 x 10^-7 = 3.397 x 10^-3 kg   OR 3.397 g

number of mol of copper in this ball = mass / molecular mass

         = 3.397 / 63.546 = 0.0535 mol

atomic number of copper is 29 so there is 29 electrons per atom.

and 1 mol contains 6.022 x 10^23 atoms per mol.

hence total number of electrons present in this ball

        = 29 x 0.0535 x 6.022 x 10^23 = 9.336 x 10^23 electrons

copper ball is charged to 90nC so number of electrins removed is n then

ne = 90 x 10^-9 C

n (1.6 x 10^-19 ) = 90 x 10^-9

n = 5.625 x 10^11 electrons removed.

fraction = (5.625 x 10^11) / (9.336 x 10^23) = 6.025 x 10^-13 ..........Ans