If a polonium-210 (Po-210) atom has just decayed, how close can the ejected

alpha particle get to a neighboring Po-210 nucleus?
Hint: W = F*r
F=kq1q2/r2

To solve this problem, we need to use Coulomb's Law, which relates the force between two charged particles to their charges and the distance between them. Coulomb's Law is given by:

F = k * q1 * q2 / r^2

Where:
- F is the force between the two charges
- k is the electrostatic constant (approximately 8.99 x 10^9 Nm^2/C^2)
- q1 and q2 are the charges of the particles
- r is the distance between the particles

In this case, we have an ejected alpha particle with a positive charge (+2e) and a neighboring Po-210 nucleus with a positive charge (+1e), where e is the elementary charge. The charge of the alpha particle is twice that of the nucleus.

Now, let's calculate the minimum distance the ejected alpha particle can get to the neighboring Po-210 nucleus. We want to find the distance r when the force F is equal to the work done W.

W = F * r

Since we know that the work done is equal to the initial potential energy of the two charges, we can write:

W = k * q1 * q2 / r

Equating W to F * r, we have:

F * r = k * q1 * q2 / r

Now we can substitute in the given charges:

k * (2e) * (1e) / r = k * (2e) * (1e) / r^2

We can cancel out k, so:

2e / r = 1e / r^2

Multiply both sides of the equation by r^2:

2e * r^2 = 1e

Divide both sides by 2e:

r^2 = 1e / 2e

r^2 = 1/2

Taking the square root of both sides:

r = √(1/2)

Simplifying:

r = √(1)/√(2)

r ≈ 0.707

Therefore, the minimum distance the ejected alpha particle can get to the neighboring Po-210 nucleus is approximately 0.707 times the distance between the centers of their nuclei.