A copper coin with a mass of m = 3 g has a positive charge Q = 1.12 μC. What fraction of b free electrons did the coin lose?

univerkov | February 13, 2021 | education

A copper coin with a mass of m = 3 g has a positive charge Q = 1.12 μC. What fraction of b free electrons did the coin lose? The density of copper is p = 8900 kg / m ^ 3, 3 g of copper contains Na = 6.02 * 10 ^ 23 atoms.

Since each copper atom has two electrons on the outer shell, the total number of free electrons is:
N = 2 * Na = 2 * 6.02 * 10 ^ 23 = 1.2 * 10 ^ 24
One electron has a charge e = 1.6 10 ^ -19 C, then the number of donated electrons is: n = Q / e = 1.12 * 10 ^ -6 / 1.6 * 10 ^ -19 = 7 * 10 ^ 12
And their share is equal to:
b = n / N = 7 * 10 ^ 12 / 1.2 * 10 ^ 24 = 5.8 * 10 ^ -12.

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