Alpha particles which have a charge +2e and mass 6.64 ✕ 10−27 kg are initially at rest and are fired directly at a stationary lead nucleus (charge +82e) at a speed of 1.95 ✕ 107 m/s. determine the distance of closest approach (in fm) before the alpha particle reverses direction. assume the lead nucleus remains stationary. assume the alpha particles are initially very far from a stationary lead nucleus.

The closest approach distance is 29.9 fm

Solution:

As per the question:

Charge on alpha-particle, q = +2e C

Charge on Lead nucleus, q' = +82e C

Speed of the ,

Mass of , M =

Now,

The kinetic energy of the :

Also the electro-static potential energy of the is given by:

where

D = distance between the nucleus of and lead.

Since, the kinetic energy of the get tranformed into electro-static potential energy, thus equating the two:

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