well, you have a + 80 e charge fired at a +1 e charge
they repel each other
How close will the proton get to the 80 protons?
when the kinetic energy of the oncoming proton is used up climbing the potential hill of the 80 protons. it stops and backs away
Ke = (1/2) m v^2
potential of test charge q' at distance r from q is
U = 9*10^9 q q' / r
q = 80 e
q' = e
so
9*10^9 (80 e^2) /r = (1/2) m v^2
e = 1.6 * 10^-19
m = proton mass = 1.7 *10^-27
v given = 3.2*10^7
so solve for r
hope r is bigger than d/2 = 7 *10^-15
so subtract 7*10^-15 from your r to get altitude from the surface
I don't know how to go about this question. We only have an initial velocity and no final velocity, distance or time.
A proton 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. The proton is fired at a speed of 3.20×107 m/s. When it passes the nucleus, how close will the proton be to the surface of the nucleus? Assume the nucleus remains at rest.
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