To what minimum distance will an alpha particle with kinetic energy `T=0.40 MeV` approach in the case of a head-on collision to
(a) a stationary `Pb` nucleus,
(b) a stationary free `Li^(7)` nucleus?

Explanation

(a) In the `pb` case we shall ignore the recoil of the nucleus both beause `pb` is quite heavy `(A_(pb)= 208= 52xxA_(HE))` as well as beause `pb` is not force. Then for head on collision, at the distance of cloest approach, the `K.E` of the `alpha`-particle must become zero (beacuse `alpha`-particle will trun back at this point). Then
`(2Z e^(2))/((4pi epsilon_(0))r_(min))=T`
(No `(4 pi epsilon_(0))` in Gaussian units.). Thus putting the values
`r_(min)= 0.591 p m`.
(b) Here we have to take account of the fact that the energy is spent in the recoil of Li necleus. Suppose `X_(1)=` coordinate of the `alpha`-particle from some arbitrary point on the line joining it to the `Li` nucleus, `x_(2)=` coordinate of `Li` nucleus with respect to the same point. Then we have the energy momentum equations
`(1)/(2)m_(1)x_(1)^(2)+(1)/(2)m_(2)x_(2)^(2)+(2xx3xxe^(2))/((4 pi epsilon_(0))|x_(1)-x_(2)|)=T`
`m_(1)x_(1)+m_(2)x_(2)=sqrt(2m_(1)T)`
Here `m_(1)=` mass of `He^(++)` nucleus, `m_(2)`= mass of `Li` nuclues. Eliminating `x_(2)`
`T+(1)/(2)m_(1)x_(1)^(2)+(1)/(2m_(2))(sqrt(2m_(1)T)-m_(1)x_(1))^(2)+(6e^(2))/((4pi epsilon_(0))(x_(1)-x_(2)))`
We complete the square on the right hand side and rewrite the above equations as
`(m_(2))/(m_(1)+m_(2))T=(1)/(2m_(2))[sqrt(m_(1)(m_(1)+m_(2)))x_(1)-sqrt((m_(1))/(m_(1)+m_(2)))sqrt(2m_(1)T)]^(2)`
`+(6e^(2))/((4pi epsilon_(0))|x_(1)-x_(2)|)`
For the least distacne of approach, the second term on the right must be greatest which implies that the first term must vanish.
Thus `|x_(1)-x_(2)|_(min)=(6e^(2))/((4 pi epsilon_(0))T)(1+(m_(1))/(m_(2)))`
Using `(m_(1))/(m_(2))=(4)/(7)` and other values we get
`|x_(1)-x_(2)|_(min)= 0.34pm`
(In Gaussian factor `4pi epsilon_(0) ` is absent).