Problem 1 - University Physics with Modern Physics

Question

Excess electrons are placed on a small lead sphere with mass 8.00 g so that its net charge is \(-3.20 \times 10^{-9} \mathrm{C}\) (a) Find the M 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 \(\mathrm{g} / \mathrm{mol}\) .

Step-by-Step Solution

Answer

(a) 2.00 × 10^10 electrons; (b) 8.60 × 10^-13 electrons per lead atom.

Step 1: Calculate the Number of Excess Electrons

The charge of an electron is approximately \(-1.60 \times 10^{-19}\, \mathrm{C}\). To find the number of excess electrons (\(M\)), we divide the net charge by the charge of a single electron:\[M = \frac{-3.20 \times 10^{-9}\, \mathrm{C}}{-1.60 \times 10^{-19} \, \mathrm{C/electron}} = 2.00 \times 10^{10} \text{ electrons}.\]

Step 2: Calculate the Number of Lead Atoms

To find the number of lead atoms, first convert the mass of the sphere from grams to moles. Use the molar mass of lead which is \(207 \, \mathrm{g/mol}\):\[\text{moles of lead} = \frac{8.00\, \mathrm{g}}{207\, \mathrm{g/mol}} \approx 0.03865 \text{ moles}.\]Then, use Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mol}\) to find the number of atoms:\[\text{number of lead atoms} = 0.03865 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 2.327 \times 10^{22} \text{ atoms}.\]

Step 3: Calculate the Excess Electrons Per Lead Atom

Finally, divide the number of excess electrons by the number of lead atoms to find the excess electrons per lead atom:\[\text{excess electrons per lead atom} = \frac{2.00 \times 10^{10} \text{ electrons}}{2.327 \times 10^{22} \text{ atoms}} \approx 8.60 \times 10^{-13} \text{ electrons per atom}.\]

Key Concepts

Lead Sphere

A lead sphere is a small object made entirely of lead, which is a heavy, dense metal known for its high atomic number and mass. Lead has many applications due to its properties, including its use in protective shields against radiation and in batteries.

• Lead, being dense, makes the sphere compact and heavy for its size.
• In this scenario, the mass of the lead sphere is given as 8.00 g.
• Lead's bulkiness results in interesting interactions with electric charge.

By understanding how electrons distribute on the surface of such a sphere, one can explore both electrostatics and atomic theory principles.

Net Charge

The net charge on an object is determined by the total amount of charge it has, due to the difference between the number of protons and electrons. In this problem, the lead sphere has a net charge of \(-3.20 \times 10^{-9} \, \, \mathrm{C}\), which means that there are more electrons than protons.

• A negative net charge indicates excess electrons.
• The net charge influences how the sphere interacts with other charged objects.

Calculating the net charge helps in determining the number of excess electrons, a concept integral to this type of physics problem.

Atomic Number

The atomic number of an element refers to the number of protons in the nucleus of its atoms, defining the element's identity. For lead, the atomic number is 82.

• This means each atom of lead has 82 protons.
• The atomic number is crucial since it also determines the element on the periodic table.

Understanding the atomic number is vital for calculating the properties of elements, such as the electron configuration and the behavior of atoms in different chemical reactions.

Atomic Mass

Atomic mass is the average mass of atoms of an element, expressed in atomic mass units (u). It accounts for all isotopes of an element, considering both their mass and abundance.

• Lead's atomic mass is approximately 207 u, as stated in this exercise.
• The atomic mass helps in converting grams to moles, a key step in the calculations involved with the problem.

This understanding allows scientists to measure the amount of substance using the mole concept and to estimate an element's properties in chemical equations.