General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 27, Problem 1E
To determine

The distance of closest approach between alpha particle and the gold nucleus.

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Each a particle in a beam of a particles has a kinetic energy of 4.5 MeV. Through what potential difference would you have to accelerate these a particles in order that they would have enough energy so that if one is fired head-on at a gold nucleus it could reach a point 1.9 10^-14 m from the center of the nucleus?
In a Rutherford scattering experiment, an a-particle (charge = +2e) heads directly toward a gold nucleus (charge = +79e). The α-particle had a kinetic energy of 5.0 MeV when very far (r→ ∞) from the nucleus. Assuming the gold nucleus to be fixed in space, determine the distance of closest approach. Hint: Use conservation of energy with PE =kq1q2/r.
Use the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron). Neutron: m = 1.67493x10-27 kg = 1.008665 u = 939.57 MeV/c² . ¹H: mH = 1.67353x10-27 kg = 1.007825 u = 938.78 MeV/c² 1 1 u = 1.6605x10-27 kg = 931.5 MeV/c² . Consider the following decay: 239 Pu 235 U+ a. 239 Pu has a mass of 239.0521634 u, 235 U has a mass of 235.0439299 u, and a has a mass of 4.002603 u. 94 92 94 92 Determine the disintegration energy (Q-value) in MeV. Q = Determine the binding energy (in MeV) for 239 Pu. 94 EB =
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