A particle with kinetic energy `T` is deflected by a spherical potential well radius `R` and depth `U_(0)` i.e., by the field in which the potential energy of the particle takes the form
`U = {(0" for "r gt R),(-U_(0)" for "r lt R):}`
where `r` is the distance from the centre of the well. Find the relationship between the aiming parameter `b` of the particle and the angle `theta` through which it deflects from initial motion direction.

Explanation

See the diagram on the next page. In the region where potential is nonzero, the kinetic energy of the partilce is, by energy conservation,
`T+U_(0)` and the momentum of the particle has the magnitude `sqrt(2m(T+U_(0)))`. On the boundary the force is radial, so the tangential component of the momentum does not change:
`sqrt(2mT)sin alpha= sqrt(2m(T+(U)_(0)) sin varphi`
so `sin varphi=sqrt((T)/(T+U_(0)) )sin alpha= (sin alpha)/(n)`
where `n=sqrt(1+(U_(0))/(T))`. We also have
`theta=2(alpha-varphi)`
Therefore
`"sin"(theta)/(2)= sin (alpha-varphi)= sin alpha cos varphi- cos alpha sin varphi`
`= sin alpha(cos varphi-(cos alpha)/(n))`
or `(n sintheta//2)/(sin alpha)=sqrt(n^(2)-sin^(2)alpha)-cos alpha`
or `((n sin theta//2)/(sin alpha)+ cos alpha)^(2)=n^(2)-sin^(2)alpha`
or `n^(2) "sin"^(2)(theta)/(2)"cot"^(2)alpha+2 n "sin"(theta)/(2) "cot" alpha+1=n^(2)"cos"^(2)(theta)/(2)`
or `cot alpha=("n cos"(theta)/(2)-1)/("n sin"(theta)/(2)`
Hence `sin alpha=("n sin"(theta)/(2))/(sqrt(1+n^(2)-2"n cos"(theta)/(2))`
Finally, the impact parameter is
`b=R sin alpha=("nR sin"(theta)/(2))/(sqrt(1+n^(2)-2"n cos" (theta)/(2)))`.