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/r
2

To determine how close the ejected alpha particle can get to a neighboring Po-210 nucleus, we need to consider the electrostatic force between them.

The electrostatic force (F) between two charged particles is given by Coulomb's Law:

F = k * q1 * q2 / r^2,

where k is the Coulomb constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.

In this case, the alpha particle has a charge of +2e (twice the elementary charge, since it consists of two protons), and the neighboring Po-210 nucleus has a charge of +2e as well.

We know that the alpha particle is ejected from the decaying Po-210 atom, so we can assume its initial distance from the neighboring Po-210 nucleus is greater than its radius (since it was inside the atom before decay).

To find the minimum distance the alpha particle can get to the neighboring Po-210 nucleus, we need to consider the work done (W) to bring the two particles closer:

W = F * r.

Since the alpha particle has a positive charge and the neighboring Po-210 nucleus also has a positive charge, the work done will be positive.

The minimum distance occurs when the work done is equal to its kinetic energy just after decay. However, we are not given information about the kinetic energy of the ejected alpha particle, so we cannot calculate its minimum distance based solely on that.

Hence, without more information about the initial kinetic energy of the alpha particle, we cannot determine the exact minimum distance it can get to the neighboring Po-210 nucleus after decay.