Question

A small piece of aluminum (atomic number 13) contains 10^15 atoms (The atomic number is the number of protons; it determines the (positive) electric charge of the nucleus and, thus, the number of electrons in a neutral atom.) If the piece of aluminum has a net positive charge of 3.0 uc what fraction of the electrons that the aluminum had when it was neutral would have had to be lost?

Answer 1

3 micro coulombs = 3 * 10E-6 coulombs charge of aluminum

13 * 10E15 * 1.6 * 10E-19 = 2.08 E-3 Coulombs - charge of atoms in Al

3 * 10E-6 / 2.08 * E-3 = 1.44 * E-3 = .00144 = .144 %

.00144 of the original electrons would have to be lost