Aproton is fired from far away toward the nucleus of a mercury atom. mercury is element number 80, and the diameter of the nucleus is 14.0 fm. part a if the proton is fired at a speed of 3.4×107 m/s , what is its closest approach to the surface of the nucleus? assume the nucleus remains at rest. express your answer to two significant figures and include the appropriate units.

The initial kinetic energy of the proton is given by:

where  is the proton mass and  is the initial speed.

As the proton approaches the nucleus, it decelerates because of the repulsive electric field and its kinetic energy converts into electric potential energy. The proton will stop at a distance r from the center of the nucleus, and its potential energy at this distance will be:

where , 80e is the charge of the nucleus of mercury (which contains 80 protons), and  is the proton charge.

For the conservation of energy,

Rewriting it, we find


This is not the final answer, however, because this is the distance reached by the proton with respect to the center of the nucleus. So, to find the distance from the surface, we should subtract the radius of the nucleus, which is half the diameter: 14/2=7 fm. So