00:01
In this question, we are given alpha particles, having kinetic energy of 7 .7 mvp being fired towards gold particle, right, or gold nucleus.
00:17
We want to use energy conservation to find what is the distance of closest approach.
00:22
So the initial total energy is just the kinetic energy of our alpha particle.
00:32
The final energy is only the potential energy of the system.
00:47
No kinetic energy.
00:48
That's the point where they are closest to each other, and the alpha particle is not moving.
00:57
Both the alpha particle and the particles are not moving.
01:00
So we just have to equate the kinetic energy to the potential energy, which in this case is the electric potential energy.
01:08
Due to the quorum interaction.
01:11
So this equals to ke times q1, q2 over r, where r is the distance between the nuclei at this particular point, which is our distance of closest approach.
01:30
Q1 and q2 is the charge of the alpha particle and the charge of the gold nuclei respectively.
01:38
So for the gold nuclei, this is given as 79e, because there are 79 protons, for the alpha particle is 2e.
01:57
Creating this to the kinetic energy, 7 .7, m .v.
02:01
We will have to convert the mep into joules to work in sir units.
02:09
So multiply by 1 .6, then power of 13.
02:12
So give us in terms of juice...