A proton is fired from far away toward the nucleus of a merc

A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element lumber 80. and the diameter of the nucleus is 14.0 fm. If the proton is fired at a speed of 3.6 x 10^7 m/s. what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

Solution

The distance of closest approach is the distance from the nucleus at which the total energy of incident particle is only potential and is given by

Mv^2 /2 = Zeze / 4pi€₀d

Where d is the distance of closest approach

Here kinetic energy K = [1.67x10^-27x(3.6x10^7)^2] /2 = 1.0822x 10^-12 J

d = Zee /4pi€₀K

d= [80x(1.6x10^-19)^2 x9x 10^9] / 1.0822x10^-12 = 1.7032 x10^-14 m = 17.03 fm

Distance of closest approach to the surface of nucleus = 17fm- radius of nucleus = 17fm- 7fm = 10fm