00:01
In this problem, we are asked to find the distance of the closest approach of the alpha particle to the center of the nucleus.
00:08
And what we're considering is a 4 .78 mega electron alpha particle from a 226 radium decay makes a head -on collision with the uranium nucleus.
00:20
The uranium nucleus has 92 protons.
00:25
We know that k1, kinetic energy 1 plus the kinetic energy 2, is equal to u1 plus u2.
00:40
That's our initial potential energy.
00:44
This is potential, and these are kinetic.
00:53
Okay, so let's do 4 .78 mega electron volts plus zero equals zero plus, and then we'll have one over 4 pi e .o.
01:11
And then this will be, get that off of there, q, qo over d.
01:23
So let's go ahead and put this on a little bit more work into this.
01:28
4 .7 .78 mega electron volts times 1 .602 times 10 to the minus 13th, over 1 mega electron bolt equals 9 .0 times 10 to the 9th newton meters squared per coolum squared times 2 times 92 times 1 .602 times 10 to the minus 19th coulums squared divided by d.
02:23
Okay so here i get d equals 9 .0 times 10 to the 9th newton meters squared per coulum times 2 times 92 times 1 .602 times 10 to the minus 19c squared over 7 .66 times 10 to the minus 13th joules...