A particle of mass m, charge `q gt 0` and initial kinetic energy K is projected from infinity towards a heavy nucleus of charge Q assumed to have a fixed position.
If the aim is perfect, how close to the center of the nucleus is the particle when it comes instantaneously to rest ?

Explanation

Fig. shows the conditions. Of the problem.
Charge on the nucleus = Ze,
Charge on the projectile = Z.e
At infinity , the angular momentum of projectile about the nucleus = mvb
When the projectile is at the minimum distance s from the nucleus, its velocity vector v. is normal to the radius vector (drawn from the centre of the nuclesu ).
`therefore` Angular momentum of the projectile about the nucleus = mv.s. From the law of conservation of angular momentum , we have,
mvb = mv.s or v. `= (vb)/(s) ` (i)
When the projectile is infinitely away from the nucleus, it has only kinetic energy (`because` P.E = 0 )given by,
`E_(K) = (1)/(2) mv^(2)`
when the projectile is at minimum distance s from the nucleus, it has both kinetic and potential energies given by,
`E_(k) = (1)/(2) mv^(2) and E_(p) = (1)/(4 pi epsilon_(0)) . ("ZZ".e^(2))/(s)`
From the law of conservation of energy, we have,
`E_(k) + E_(p) = E_(k)`
or `(1)/(2) mv^(2) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2))/(s) = (1)/(2) mv^(2)`
putting v. = vb/s from eq. (i), we get,
`(1)/(2) m (v^(2) b^(2))/(s^(2)) + (1)/( 4pi epsilon_(0)) .("ZZ".e^(2))/(s) = (1)/(2) mv^(2)`
Dividing both sides by `mv^(2) `/2, we get,
`(b^(2))/(s^(2)) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2))/(s(mv^(2)/2)) = " or " s^(2) = b^(2) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)`
This is the required formula.
For a head- on collision , b = 0 .
`therefore s^(2) = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2) " or " therefore s = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)`
Which is the distance of closest approach