Problem 23.89 An alpha particle with kinetic energy 13.5MeV makes a collision wi

Question

Problem 23.89 An alpha particle with kinetic energy 13.5MeV 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=1.50x10^-12m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead ?s 82. The alpha particle is a helium nucleus, with atomic number 2.) Part A What is the distance of closest approach? Part B Repeat for b=1.50*10^-13 m.

Explanation / Answer

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

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