00:01
According to the question we are given that coolant potential is represented by br that is equals to z1, z2 into e squared divided by r.
00:13
Here z1 and z2 represents the charges, the charges of projectile, projectile and target particles, target particles, while r represents the radiae between them.
00:32
Or the radius between them.
00:34
We need to find the differential cross -section formula.
00:39
Differential cross -section using the formula.
00:44
We know that to find this, the formula is that f -theta is equals to minus 2 -new divided by qh -square into the integral from 0 to infinity for vr, sine qr, r, dr, this represents our equation.
01:03
Number 1.
01:04
Now put this value of vr in this equation.
01:09
Therefore we have v -the -teta will be equal to.
01:13
We know that a will be equal to 1 upon r0.
01:17
We have to use this only.
01:19
Therefore f -the - is equal to minus 2 -new upon uh square integral from 0 to infinity of z1, z2, e -square upon r into e minus r -n -nus, into r, sine qr, dr, this becomes a equation...