Asked by Marty

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
Answered by bobpursley
The particle has initial KE, lets say K.

That energy can be converted to PE as it approaches the nucleus.

When equal, the particle can go no further.

K= INTEGRAL F*dr= -kq1q2/r evaluated from infinity to x

K= kq1q2/x

so the minimum distance x= kq1q2/K

So look up the energy lost when the Po nucleus decays, the charge on the decayed nucleus, and the charge on the alpha particle. Then, it is just calculator work.
Answered by Marty
I don't quite understand
Answered by drwls
According to my physics text, Po-210 decays by alpha decay, leaving an alpha particle (+2e charge) , a Pb-206(-2) ion that will eventually neautalize, and 45.17 - 36.15 = 9.02 MeV of energy.

Most, but not all, of the energy will be kinetic energy of the alpha particle. The adjacent Po atom has nuclear charge of +84 e.

They probably expect you to ignore the energy loss as the alpha particle penetrates the electron cloud around the Po atom.
Answered by Bobby
Where does the 45.17-36.15 come from and the +84 e?
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