Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is – 3.20 X 10-9 C.
(a) Find the number of excess electrons on the sphere.
(b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol.
The Correct Answer and Explanation is :
Problem Analysis
We are tasked with finding:
- The number of excess electrons on a lead sphere with a known mass and net charge.
- The number of excess electrons per lead atom, considering the atomic properties of lead.
Part (a): Number of Excess Electrons
The charge of a single electron is e=−1.60×10−19 Ce = -1.60 \times 10^{-19} \, \text{C}.
The sphere’s net charge is Q=−3.20×10−9 CQ = -3.20 \times 10^{-9} \, \text{C}.
The number of excess electrons, NeN_e, is given by: Ne=QeN_e = \frac{Q}{e}
Substituting the values: Ne=−3.20×10−9−1.60×10−19=2.00×1010N_e = \frac{-3.20 \times 10^{-9}}{-1.60 \times 10^{-19}} = 2.00 \times 10^{10}
Thus, the sphere has 2.00×10102.00 \times 10^{10} excess electrons.
Part (b): Excess Electrons per Lead Atom
Step 1: Calculate the number of lead atoms in the sphere
The sphere’s mass is m=8.00 g=0.00800 kgm = 8.00 \, \text{g} = 0.00800 \, \text{kg}.
The molar mass of lead is M=207 g/molM = 207 \, \text{g/mol}, and 1 mole contains NA=6.022×1023N_A = 6.022 \times 10^{23} atoms.
The number of lead atoms, NatomsN_\text{atoms}, is: Natoms=mass of spheremolar mass×NAN_\text{atoms} = \frac{\text{mass of sphere}}{\text{molar mass}} \times N_A
Substituting values: Natoms=8.00207×6.022×1023=2.33×1022 atomsN_\text{atoms} = \frac{8.00}{207} \times 6.022 \times 10^{23} = 2.33 \times 10^{22} \, \text{atoms}
Step 2: Excess electrons per lead atom
The excess electrons per atom, Neper atomN_e^\text{per atom}, is: Neper atom=NeNatomsN_e^\text{per atom} = \frac{N_e}{N_\text{atoms}}
Substituting values: Neper atom=2.00×10102.33×1022=8.58×10−13N_e^\text{per atom} = \frac{2.00 \times 10^{10}}{2.33 \times 10^{22}} = 8.58 \times 10^{-13}
Explanation
The lead sphere’s negative charge results from 2.00×10102.00 \times 10^{10} excess electrons. Each electron contributes a charge of −1.60×10−19 C-1.60 \times 10^{-19} \, \text{C}, and dividing the total charge by this value gives the number of electrons.
To determine the distribution of these electrons, we first calculate the number of lead atoms using the sphere’s mass, the molar mass of lead, and Avogadro’s number. The result, 2.33×10222.33 \times 10^{22} atoms, indicates the immense number of atoms present. Dividing the total excess electrons by the number of atoms gives 8.58×10−138.58 \times 10^{-13} excess electrons per atom, a negligible amount compared to the number of electrons naturally present in each lead atom.
Thus, the additional electrons slightly alter the charge distribution without significantly impacting the sphere’s atomic structure.