Answer:
[tex]f = 2.6 \times 10^{-13}[/tex]
Explanation:
Let the mass of copper ball is "m" gram
now the total number of copper atom present in the ball is given as
[tex]N = \frac{m}{29} \times 6.02 \times 10^{23}[/tex]
now the total number of electrons in one copper atom is 29
so total number of electrons in given sample of copper ball is
[tex]N_e = m(6.02 \times 10^{29})[/tex]
now diameter of the ball is 7.0 mm
density of the ball = [tex]8900 kg/m^3[/tex]
now we have
[tex]m = (\frac{4}{3}\pi r^3)(8900)[/tex]
[tex]m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)[/tex]
[tex]m = 1.6 gram[/tex]
now we have
[tex]N_e = 9.63 \times 10^{23}[/tex]
now the charge on the copper ball is 40 nC
so the number of electrons removed
[tex]Q = ne[/tex]
[tex]40 \times 10^{-9} = n(1.6 \times 10^{-19}[/tex]
[tex]n = 2.5 \times 10^{11}[/tex]
so the fraction of number of electrons removed is given as
[tex]f = \frac{n}{N_e}[/tex]
[tex]f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}[/tex]
[tex]f = 2.6 \times 10^{-13}[/tex]
