Nuclear Fission Problem solved correctly? Need a check.?

Question

In nuclear fission, a nucleus of uranium -238, which contains 92 protons, divides into smaller spheres, each having 46 protons and a raduis of 5.9 x 10^-15 m. What is the net force on this pair of molecules?

using
kq1q2/r2
i got
(8.99 x 10^9) (46*(1.6x10^-19))^2
----------------------- -----------------------
(2*(5.9x10^-15))^2

= 3.47 x 10^23 N

Answer 1

Everything looks good, except your final answer is way off.

Checking exponents
9 + (2*-19) - 2* (-15)

= 9-38+30=1. So your answer should be a couple orders of magntude greater than that. But not 10^23! Try plugging the numbers in again--you should get something in the range of thousands of newtons.

Also, it's a bit naive to think that electrostatic repulsion is the only thing going on here. But for a homework problem, we'll let that slide.

Answer 2

You clearly made this question up. It has no answer because it has no basis in science.

U238 does not necessarily split into equal halves when fission occurs. In fact, it may split into more than two parts. Second, the fissioned parts are not a "pair of molecules," they are still atoms. Third, the fissioned parts are not bound to each other in any way; so there is no "net force" between them.

Perhaps what you are looking for is the energy produced when U238 fissions. That comes from good old E = mc^2. In fact, the REAL equation is E = delm c^2, where delm is the difference (del) between the U238 mass m0 before fissioning and the sum of the masses (Sum m) of the parts after fissioning. That is, del m = m0 - Sum m.

What we find is that Sum m < m0; that is, some of the mass m0 is lost so the sum of the masses after splitting is less than what the U238 started with. What happened to that lost mass...it became energy...E = delm c^2. When billions and billions of U238 atoms do this, we have nuclear power plants or, if the splitting happens more rapidly...the atomic bomb.

Answer 3

The force would be (46*e)^2 / 4πe0*(2r)^2, where e is the electronic charge 1.6*10^-19 coul, e0 = permittivity of space, 8.85*10^-12 farad/m.

I get 3.5*10^3 N

Answer 4

I'm sorry but the correct answer is

(8.99 x 10^9) (46*(1.6x10^-19))^2
----------------------- -----------------------
(2*(5.9x10^-15))^2

= 3.47 x 10^23 N +2

Answer 5

yes



\