# physics

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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

• physics -

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.

• physics -

I don't quite understand

• physics -

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.

• physics -

Where does the 45.17-36.15 come from and the +84 e?