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=1.30×10^12 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? b) Repeat for b=1.40x10^-13 m c) Repeat for b=1.50x10^-14 m

Explanation / Answer

In this problem we need to use energy conservation. The initial energy (kinetic energy of alpha particle) should be equal to the final energy, which is the potential energy of interaction of alpha particle and a lead nucleus. Then

Einitial = Efinal = Ke*(Qalpha*Qlead) / r

We know the initial energy:

Einitial = 13 MeV = 13 x 10^6 x 1.6 x 10^-19 = 2.08 x 10^-12 C

We know the charge of the particles:

Qalpha = 2e = 2 x 1.6 x 10^-19 = 3.2 x 10^-19 C

Qlead = 82e = 82 x 1.6 x 10^-19 = 131.2 x 10^-19 C

Then we can find the shortest distance

r = Ke*(Qalpha*Qlead) / Einitial

r = (9 x 10^9 x 3.2 x 10^-19 x 131.2 x 10^-19) / (2.08 x 10^-12)

r = 0.1816 x 10^-13 m

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