An alpha particle with kinetic energy `T= 0.50 MeV` is deflected through an angle of `theta=90^(@)` by Coulomb field of a stationary `Hg` nucleus. Find:
(a) the least curvature radius of its trajectory,
(b) the minimum approach distance between the particle and the nucleus.

Explanation

We shall ignore the recoil of `Hg` nucleus.
(a) Let `A` be the point of closest approach to the centre `C, AC= r_(min)`. At `A` the motion is instantaneously circular because the radial velocity vanishes. Then if `v_(0)` is the speed of the particle at `A`, the following equation hold
`T=(1)/(2)mv_(0)^(2)+(Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0))r_(min))` (1)
`mv_(0)r_(min)= sqrt(2mT)b` (2)
`(mv_(0)^(2))/(rho_(min))=(Ƶ_(1)Ƶ_(2)e^(2))/((4 pi epsilon_(0))r_(min)^(2))` (3)
(This is Newton's law. Here `rho=rho_(min)` is the radius curvature of the path at `A` and `rho` is minimum at `A` by symmetry.) Finally we have Eqn. (6.1 a) in the form
`b=(Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0))2T)"cot"(theta)/(2)` (4)
From (1) and (3) `(2Tb^(2))/(rho_(min))= (Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0)))`
or `rho_(min)=(Ƶ_(1)Ƶ_(2)e^(2))/((4 pi epsilon_(0))2T)"cot"^(2) (theta)/(2)`, with `z_(1)= 2,z_(2)= 80` we get
`rho_(min)= 0.231 p m`
(b) From (2) and (4) we write
`r_(min)=(Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0))sqrt(2mT))(cot theta//2)/(v_(0))`
Substituting in (i) `T= (1)/(2)mv_(0)^(2)+sqrt(2mT)v_(0)tan theta//2`
Solviing for `v_(0)` we get `v_(0)sqrt((2T)/(m))("sec"(theta)/(2)-"tan"(theta)/(2))` Then `r_(min)=(Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0))2T)("cot"(theta)/2)/("sec"(theta)/(2)-"tan"(theta)/(2))`
`=(Ƶ_(1)Ƶ_(2)e^(2))/((4 pi epsilon_(0))2T)"cot"(theta)/(2)("sec"(theta)/(2)+"tan"(theta)/(2))`
`=(Ƶ_(1)Ƶ_(2)e^(2))/((4pi epsilon_(0))2T)(1+"cosec"(theta)/(2))=0.557` pm