A 3.0-g copper penny has a net positive charge of \({\bf{32}}\;{\bf{\mu C}}\). What fraction of its electrons has it lost?

Electric charge is the property of an elementary particle which gives rise to the electric force between the two elementary particles.

It is of two kinds, positive and negative. The value of the basic unit of charge present on an electron or proton is \(e = 1.602 \times {10^{ - 19}}\;{\rm{C}}\).

The total charge on a body is always quantized, i.e., it is the integral multiple of the basic unit of charge.

\(q = ne\)

Here, n is an integer.

Mass of the copper penny is, \(M = 3.0\;{\rm{g}} = 3.0 \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}\)

The net positive charge on the copper penny is, \(Q = 32\;{\rm{\mu C}} = 32 \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{C}}\)

The atomic number of copper is 29.

The molecular weight of copper is,\(m = 63.5\;{\rm{g}} = 63.5 \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}\)

The magnitude of the charge on each electron is, \(e = 1.602 \times {10^{ - 19}}\;{\rm{C}}\)

One atom of copper contains 29 electrons. Thus, the total charge of electrons in each atom of copper is,

\(\begin{aligned}{c}q &= 29e\\ &= 29 \times 1.602 \times {10^{ - 19}}\;{\rm{C}}\\ &= 46.46 \times {\rm{1}}{{\rm{0}}^{ - 19}}\;{\rm{C}}\end{aligned}\)

The number of atoms contained in one mole of copper is \(6.02 \times {10^{23}}\). Since the molecular weight of copper is\(63.5 \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}\). Thus, you can say that there are \(6.02 \times {10^{23}}\;{\rm{atoms}}\) in \(63.5 \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}\).

So, 1 kg of copper contains \(\frac{{6.02 \times {{10}^{23}}}}{{63.5 \times {\rm{1}}{{\rm{0}}^{ - 3}}}}\;{\rm{atoms}}\).

The number of atoms in a copper penny is,

\(\begin{aligned}{c}n &= \left( {3.0 \times {\rm{1}}{{\rm{0}}^{ - 3}}} \right) \times \frac{{6.02 \times {{10}^{23}}}}{{63.5 \times {\rm{1}}{{\rm{0}}^{ - 3}}}}\;{\rm{atoms}}\\ &= 0.284 \times {10^{23}}\;{\rm{atoms}}\end{aligned}\)

The total amount of charge in copper penny due to electrons is equal to the number of atoms of copper times the charge contained in each atom, i.e.,

\(\begin{aligned}{c}Q' &= nq\\ &= 0.284 \times {10^{23}} \times 46.46 \times {\rm{1}}{{\rm{0}}^{ - 19}}\;{\rm{C}}\\ &= 1.3{\rm{2}} \times {\rm{1}}{{\rm{0}}^5}\;{\rm{C}}\end{aligned}\)

Since copper penny has Q amount of net positive charge, that means the total charge on electrons which are lost is Q.

So, the fraction of electrons that has been lost is:

\(\begin{aligned}{c}\frac{Q}{{Q'}} &= \frac{{32 \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{C}}}}{{1.3{\rm{2}} \times {\rm{1}}{{\rm{0}}^5}\;{\rm{C}}}}\\ &= 24.24 \times {\rm{1}}{{\rm{0}}^{ - 11}}\\ &= 2.4 \times {\rm{1}}{{\rm{0}}^{ - 10}}\end{aligned}\)

Thus, the fraction of electrons lost by a copper penny is \(2.4 \times {\rm{1}}{{\rm{0}}^{ - 10}}\).