Question

An alpha particle with kinetic energy 15.0MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p0b, where p0 is the magnitude of the initial momentum of the alpha particle and b=1.30

Answer 1

The total energy at closest approach is the kinetic energy + potential energy at that point. Let v0 be the velocity and r0 the distance at the closest approach. The total energy there is then
PE = K*q1*q2/r0
KE = 0.5*m*v0^2

Total energy U(r0) = K*q1*q2/r0 + 0.5*m*v0^2

Using conservation of angular momentum,

initial = v1*m*b
at r0 = v0*m*r0

these must be equal so v1*m*b = v0*m*r0

v0 = v1*b/r0

substitute this for v0 in the energy equation

U(r0) = K*q1*q2/r0 + 0.5*m*v1^2*b^2/r0^2

this must equal the initial kinetic energy U(1) (15 MeV)

U1 = K*q1*q2/r0 + 0.5*m*v1^2*b^2/r0^2

however, 0.5*m*v1^2 = U1

U1 = K*q1*q2/r0 + U1*b^2/r0^2

U1*r0^2 - (K*q1*q2)*r0 - U1*b^2 = 0

r0^2 - (K*q1*q2/U1)*r0 - b^2 = 0

U1 = 2.40*10^-12 J = 15MeV
K*q1*q2 = 3.789*10^-26
K*q1*q2/U1 = 1.5787*10^-14

r0^2 - 1.5787*10^-14*r0 - 1.69*10^-26 = 0

r0 = 1.38*10^-13 m

B. when b = 1.5*10^-14

r0^2 - 1.5787*10^-14*r0 - 2.25*10^-28 = 0

r0 = 2.48*10^-14