Problem 8 - University Physics with Modern Physics

Question

Two small aluminum spheres, each having mass 0.0250 \(\mathrm{kg}\) are separated by \(80.0 \mathrm{cm} .\) (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 \(\mathrm{g} / \mathrm{mol}\) , and its atomic number is \(13 . )\) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{N}\) (roughly 1 ton \() ?\) Assume that the spheres may be treated as point charges. (c) What fraction of all the electrons in each sphere does this represent?

Step-by-Step Solution

Answer

(a) Each sphere contains approximately \( 7.25 \times 10^{24} \) electrons. (b) Transfer approximately \( 1.49 \times 10^{15} \) electrons. (c) This is about \( 2.06 \times 10^{-10} \) of the electrons in a sphere.

Step 1: Calculate the number of moles in a sphere

To find the number of moles, determine how much mass of aluminum you have compared to its molar mass. Use the formula: \[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \]For each sphere, we have: \[ \text{Moles} = \frac{0.0250 \text{ kg}}{26.982 \text{ g/mol}} = \frac{25.0 \text{ g}}{26.982 \text{ g/mol}} \approx 0.926 \text{ moles} \]

Step 2: Calculate the number of aluminum atoms per sphere

Use Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ atoms/mol} \) to find the number of atoms:\[ \text{Number of atoms} = 0.926 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol}\]This results in approximately:\[ 5.578 \times 10^{23} \text{ atoms} \]

Step 3: Calculate the number of electrons per sphere

Each aluminum atom has 13 electrons (atomic number 13). Thus, the total number of electrons is:\[ \text{Number of electrons/sphere} = 5.578 \times 10^{23} \text{ atoms} \times 13 \approx 7.25 \times 10^{24} \text{ electrons} \]

Step 4: Calculate the number of electrons to transfer

Use Coulomb's Law for the force between two point charges: \[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \] Solve for the amount of charge. Given \( F = 1.00 \times 10^{4} \text{ N} \), \( r = 0.8 \text{ m} \), \( k = 8.99 \times 10^{9} \text{ N m}^2/ \text{C}^2 \), and knowing \( q_1 = q_2 = q \), solve:\[ q^2 = \frac{F \cdot r^2}{k} \]\[ q = \sqrt{\frac{1.00 \times 10^{4} \cdot (0.8)^2}{8.99 \times 10^{9}}} \approx 2.39 \times 10^{-4} \text{ C} \]\[ n = \frac{q}{e} \approx \frac{2.39 \times 10^{-4} \text{ C}}{1.60 \times 10^{-19} \text{ C/electron}} \approx 1.49 \times 10^{15} \text{ electrons} \]

Step 5: Calculate the fraction of electrons transferred

Compute the fraction of electrons transferred relative to the total number of electrons in a sphere:\[ \text{Fraction} = \frac{1.49 \times 10^{15}}{7.25 \times 10^{24}} \]This results in:\[ \text{Fraction} \approx 2.06 \times 10^{-10} \]

Key Concepts

Atomic Structure

Atoms are the basic building blocks of matter, composed of protons, neutrons, and electrons. Understanding atomic structure is essential for grasping how materials behave and interact. In our exercise, each aluminum sphere is composed of aluminum atoms.

• Protons and neutrons make up the nucleus of an atom, which is positively charged due to protons.
• Electrons, which are negatively charged, orbit the nucleus in electron "clouds" or shells.
• An aluminum atom contains 13 protons and 13 electrons, giving it a neutral charge under normal conditions.

Knowing that each sphere contains a large number of atoms helps us understand the large number of electrons present in each sphere. These electrons can be transferred to create electrical forces, as seen in Coulomb's Law interactions.

Electron Transfer

Electron transfer is the movement of electrons from one atom or molecule to another. This process is necessary for creating the charge imbalance that results in electrical forces. In the context of our exercise, electrons are transferred from one aluminum sphere to another.

• Each aluminum atom has 13 electrons.
• Transferring electrons between spheres creates charges, turning each sphere into a charged body.
• This charge difference is what creates attractive or repulsive forces between the spheres.

The amount of electron transfer necessary to create a measurable force, such as in Coulomb's Law, is calculated by considering factors such as the desired force and distance between charges. In this exercise, Coulomb's Law quantifies the number of electrons needed to be transferred to achieve a force of 10,000 Newtons.

Point Charges

Point charges are idealized charges that allow us to simplify calculations in physics by assuming the charge is concentrated at a single point in space. For exercises like this one, we assume each aluminum sphere acts as a point charge.

• Point charges simplify the application of Coulomb's Law, which uses the distance between charges to calculate force.
• This model doesn't account for the physical size of the objects, just their charge and separation distance.
• Using point charges is a valid approximation when the spheres are far enough apart relative to their size.

This simplification allows for straightforward calculations of the forces between charged objects, such as spheres in our problem. Assuming point charges allow us to focus on understanding the principles of electrical interaction.

Molar Mass Calculation

Calculating molar mass is a key step in converting the mass of a substance to the number of moles, which is crucial in chemical calculations as shown in our exercise.

• Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol).
• To calculate moles, the formula is: \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \).
• For aluminum, with a molar mass of 26.982 g/mol, calculating the moles helps determine the number of atoms present in the spheres.

By knowing the number of moles, we can use Avogadro's number \( (6.022 \times 10^{23} \text{ atoms/mol}) \) to find the total atoms and, consequently, the number of electrons in the spheres. This information is crucial for calculating electron transfer and electrical forces.