An alpha particle with kinetic energy 13.0 MeV makes a collision with lead nucle

Question

An alpha particle with kinetic energy 13.0 MeV 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=1.10×1012 m . (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

A. What is the distance of closest approach? (I tried doing this and I got 1.01x10^-12 m and that's wrong, also 0 is not the answer)

B. Repeat for b=1.2x10^-13 m

C. Repeat for b=1.3x10^-14 m

Explanation / Answer

rmin = bcosQ/2/(1-sinQ/2)

when Q is the deflection angle and can be found by

b = zke^2/Ek *sqrt[(1+cosQ)/(1-cosQ)]

(1+cosQ)/(1-cosQ) = (bEk/zke^2)^2 = (1100 fm*13.0MeV/(82*1.44MeV fm))^2 =14666.26

1+cosQ = 14666.26 -14666.26cosQ

cosQ = 14665.26/14667.26 = 0.999864

Q = 0.9462 degrees

rmin = bcosQ/2/(1-sinQ/2) = 1.11x10^-12 m

for b = 1.2x10^-13, (1+cosQ)/(1-cosQ) = (120*13.0/(82*1.44))^2 = 174.54, cosQ = 173.54/175.54 = 0.9886, Q = 8.657 degrees, rmin = 1.294x10^-13 m

for c, = 1.3 x10^-14, (1+cosQ)/(1-cosQ) = (13*13/82/1.44)^2 = 2.048, cosQ = 1.048/3.0248 = 0.3438, Q =69.89, rmin = 1.3x10^-14*cos34.94/(1-sin34.94) = 2.494 x10^-14 m

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