COLLEGE PHYSICS,V.2
11th Edition
ISBN: 9781305965522
Author: SERWAY
Publisher: CENGAGE L
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Chapter 28, Problem 6P
To determine
The distance of closest approach.
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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.
In a Rutherford scattering experiment, an ?-particle (charge = +2e) heads directly toward a silver nucleus (charge = +47e). The ?-particle had a kinetic energy of 4.6 MeV when very far (r → ∞) from the nucleus. Assuming the silver nucleus to be fixed in space, determine the distance of closest approach (in fm). Hint: Use conservation of energy with PE =(keq1q2 / r)
_____fm
Alpha particles are projected toward a gold foil from a distance that is sufficiently large to consider the Coulomb force negligible. The gold nuclei have 118 neutrons and 79 protons. If a 3.60 MeV alpha particle has a scattering angle of 180° and the gold nucleus does not recoil, determine the distance of closest approach of the alpha particle.
Chapter 28 Solutions
COLLEGE PHYSICS,V.2
Ch. 28.3 - Prob. 28.1QQCh. 28.4 - Prob. 28.2QQCh. 28.5 - Prob. 28.3QQCh. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQ
Ch. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQCh. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46APCh. 28 - Prob. 47APCh. 28 - Prob. 48APCh. 28 - Prob. 49APCh. 28 - Prob. 50APCh. 28 - Prob. 51APCh. 28 - Prob. 52APCh. 28 - Prob. 53APCh. 28 - Prob. 54AP
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- (a) Show that if you assume the average nucleus is spherical with a radius r=r0A1/3, and with a mass at A u, then its density is independent at A. (b) Calculate that density in u/fm3 and kg/m3, and compare your results with those found in Example 31.1 for 56Fe.arrow_forward(a) An aspiring physicist wants to build a scale model of a hydrogen atom for her science fair project. If the atom is 1.00 m in diameter, how big should she try to make the nucleus? (b) How easy will this be to do?arrow_forwardIn a Rutherford scattering experiment, assume that an incident alpha particle (radius 1.80 fm) is headed directly toward a target gold nucleus (radius 6.23 fm).What energy must the alpha particle have to just barely “touch” the gold nucleus?arrow_forward
- In a Rutherford scattering experiment, an a - particle (charge = 12e) heads directly toward a gold nucleus (charge = +79e). The alpha - particle had a kinetic energy of 5.0 MeV when very far (r -> infinity`) 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 = ke q1q2/r.arrow_forwardWhen an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a 5.00 MeV alpha particle has a headon elastic collision with a gold nucleus that is initially at rest.What is the kinetic energy of (a) the recoiling nucleus and (b) the rebounding alpha particle?arrow_forwardUse the below values for this problem. Please note that the mass for H is for the entire atom (proton & electron). Neutron: m,= 1.67493x1027 kg= 1.008665 u = 939.57 MeVIC H: my = 1.67353x10 27 kg = 1.007825 u = 938.78 MeVic 1u= 1.6605x10-27 kg = 931.5 MeVic? Consider the following decay: 211 At 207 Bi + a. 211 At has a mass of 210.9874963 u, 207 Bi has a mass of 206.981593 u, and a has a mass of 4.002603 u. 85 83 85 83 Determine the disintegration energy (Q-value) in MeV. Determine the binding energy (in MeV) for 211 At. 85 EB =arrow_forward
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