Question

Two small aluminum spheres, each having mass 0.0250 kg, are separated by 80.0 cm. (a) how many electrons does each sphere contain? (the atomic mass of aluminum is 26.982 g>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 * 104 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?

Answer 1

Part (a):
number of electrons is represented by the atomic number.
So, in order to solve this part, we will get the number of atoms and multiply it by the atomic number to get the number of electrons.
number of moles = mass/molar mass = 0.0250/0.026982 = 0.926 moles 
number of atoms = number of moles x Avogadro's number
                            = 0.926 x 6.022 x 10^23 = 5.579 x 10^23 atoms
number of electrons = number of atoms x atomic number
                                = 5.579 x 10^23 x 13 = 7.25 x 10^24 electrons

Part (b):
Coulomb's law can be written as follows:
F = (k x q1 x q2) / r^2
We will assume that q1 is equal to q2 in magnitude but opposite in direction
Therefore, q1 = q and q2 = -2
F = (-k x q^2) / r^2 
Substitute in the equation by the force, coulomb's constant and the radius to get q and then divide q with the fundamental charge to get number of electrons as follows:
1.00 x 10^4 = -(8.9 x 10^9 x q^2)/(0.8)^2
q = 8.479 x 10^-4 (ignoring the negative sign as it states only direction)
number of electrons = 1.6 x 10^-29 x 8.479 x 10^-4  = 1.35 x 10^-22 electron

Part (c):
To get the fraction, simply divide your answer from part (b) by your answer from part (a) as follows:
fraction = (1.35 x 10^-22) / (7.25 x 10^24) = 1.86 x 10^-47

Answer 2

(a)

m = mass of each aluminium sphere = 0.0250 kg = 0.0250 x 1000 g = 25 g

M = atomic mass = 26.982 g/mol

n = number of moles

Number of moles is given as

n = m/M

n = 25/26.982

n = 0.93

N₀ = Avogadro's number = 6.023 x 10²³

N = number of atoms

Number of atoms is given as

N = n N₀

N = (0.93) (6.023 x 10²³) = 5.6 x 10²³ atoms

$n_{e}$ = number of electrons per atom = 13

$N_{e}$ = total number of electrons

Total number of electrons are given as

$N_{e}$ = $n_{e}$ N

$N_{e}$ = (13) (5.6 x 10²³)

$N_{e}$  = 72.8 x 10²³ electrons

b)

F = magnitude of electric force between the two spheres = 1 x 10⁴ N

r = distance between the spheres = 80 cm = 0.80 m

Q = magnitude of charge on each sphere = ?

Using Coulomb's law, force between the spheres is given as

F = k Q²/r²

inserting the values

1 x 10⁴ = (9 x 10⁹) Q²/(0.80)²

Q = 8.4 x 10⁻⁴ C

n = number of electrons removed

e = magnitude of charge on each electron

using the equation

Q = n e

8.4 x 10⁻⁴ = n (1.6 x 10⁻¹⁹)

n = 5.25 x 10¹⁵

c)

Fraction = n/$N_{e}$  = (5.25 x 10¹⁵)/(72.8 x 10²³) = 7.2 x 10⁻¹⁰