An aluminum nucleus has 13 protons and 14 neutrons. A copper nucleus has 29 protons and 34 neutrons. What is the force of repulsion between an aluminum nucleus and a copper nucleus when they are separated by 10-9 m?

To calculate the force of repulsion between two nuclei, we can use Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:

F = k * (q1 * q2) / r^2

Where:
F is the force of repulsion,
k is the Coulomb's constant (approximately equal to 9 x 10^9 N*m^2/C^2),
q1 and q2 are the charges of the two nuclei,
and r is the distance between the two nuclei.

In this case, the charges of the nuclei can be considered equal to their number of protons since protons carry positive charge.

For the aluminum nucleus (13 protons), q1 = 13e (since the elementary charge, e, is equal to 1.6 x 10^-19 C).

For the copper nucleus (29 protons), q2 = 29e.

The distance between them, r, is given as 10^-9 m.

Now, plugging in these values into the formula:

F = (9 x 10^9 N*m^2/C^2) * [(13e) * (29e)] / (10^-9 m)^2

Calculating the product of the charges:

q1 * q2 = (13e) * (29e) = 377e^2

Substituting back into the formula:

F = (9 x 10^9 N*m^2/C^2) * (377e^2) / (10^-9 m)^2

Simplifying further:

F = (9 x 10^9 N*m^2/C^2) * (377) / (10^-9 m)^2 * (e^2)

Evaluating the known values:

F ≈ 1.353 x 10^-3 N

Therefore, the force of repulsion between the aluminum nucleus and the copper nucleus, when separated by 10^-9 m, is approximately 1.353 x 10^-3 Newtons.