Problem 23.89 Part A What is the distance of closest approach? An alpha particle

Question

Problem 23.89 Part A What is the distance of closest approach? An alpha particle with kinetic energy 15.0 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 Pob, where po is the magnitude of the initial 0? Im momentum of the alpha particle and b 1.10x10-12 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.) Submit My Answers Give Up Part B Repeat for bel .50×10-13 m t1l Submit My Answers Give Up Part C Repeat for b 1.00x10-14 m 17 tm

Explanation / Answer

Given,

The energy of the alpha particle K = 15.0 MeV =15 x (1.6 x 10-13) J

Charge on the electron, q = 1.6 x 10-19 C

k = 9 x 109 N-m2/C2

Atomic number of Lead, Z = 82

From law of conservation of energy, kinetic energy = Potential energy

K = k ( Zq)( 2q)/ d

So, closest distance, d =[k(Zq)( 2q)]/( K) = [2 Z k q2]/(K)

= [2 x 82 x 9 x 109 x (1.6 x 10-19)2]/(15 x 1.6 x 10-13) = 1.57 x 10-14 m

Repeat for other values... Please rate...

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