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
I am not sure where the 45.17-36.15 = 9.02 MeV of energy comes from and the +84 e comes from?
To determine how close the ejected alpha particle can get to a neighboring Po-210 nucleus after it decays, we need to consider the electrostatic force between the two particles.
The electrostatic force between two charged particles can be calculated using Coulomb's law:
F = k * (q1 * q2) / r^2
Where:
- F is the electrostatic force between the two particles
- k is the electrostatic constant (k ≈ 9 × 10^9 N·m^2/C^2)
- q1 and q2 are the charges of the two particles
- r is the distance between the particles
In this case, both the alpha particle and the Po-210 nucleus have positive charges, so we can assume they repel each other. Therefore, the force is repulsive.
Now, since we are asked to determine the distance at which the alpha particle can get to the neighboring Po-210 nucleus, we need to look at the work done by the electrostatic force.
The work done (W) is equal to the force (F) multiplied by the distance (r):
W = F * r
Now we can substitute the expression for the force into the work equation and solve for the distance (r):
W = (k * q1 * q2) / r
Rearranging the equation to isolate r, we have:
r = (k * q1 * q2) / W
Given the charge of the alpha particle (q1) and the Po-210 nucleus (q2) and the work done by the force, you can calculate the distance (r) at which the ejected alpha particle can get to the neighboring Po-210 nucleus after it decays.