Question

In a student lab experiment, 4.8 MeVMeV alpha particles from the decay of 210Po210Po are directed at a piece of thin platinum foil.

If an alpha particle is directed straight toward the nucleus of a platinum atom, what is the distance of closest approach?

Answer 1

At the distance of closest approach ,kinetic energy of alpha article is converted into electrostatic potential energy of system of alpha particle and platinum nucleus. (By conservation of energy)

That is ............................(1)

Where

=kinetic energy of alpha=4.8Mev=4.8xx1.6xjoule

=charge of alpha=2e=2xC

=charge of platinum nucleus=78e=78xC (atomic number of platinum=78)

r=distance of closest approach=?

From equation (1),

=9x

r=9xx2xx78x/[4.8xx1.6x] =1.828xmeter=distance of closest approach