College Physics: A Strategic Approach (3rd Edition)
3rd Edition
ISBN: 9780321879721
Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher: PEARSON
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Chapter 29, Problem 52GP
To determine
The name of the element.
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When an alpha particle collides elastically with a nucleus, the nucleus recoils. Suppose a 6.00 MeV alpha particle has a head-on elastic collision with a silver-108 nucleus that is initially at rest. (a) What is the kinetic energy of the recoiling nucleus? MeV (b) What is the kinetic energy of the rebounding alpha particle? MeV
An alpha particle with kinetic energy 11.0 Me V
makes a collision with lead nucleus, but it is not
"aimed" at the center of the lead nucleus, and has
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L%=pob, where po is the magnitude of the initial
momentum of the alpha particle and
b=1.50x10-12m (Assume that the lead nucleus
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point charge. The atomic number of lead is 82. The
alpha particle is a helium nucleus, with atomic
number 2.)
Repeat for b=1. 10×10-13 m.
Express your answer in meters.
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Part C
Repeat for b=1.50×10-14 m.
Express your answer in meters.
In Rutherford’s original experiment, alpha particles (α) were shot toward a gold foil. Gold has an atomic number of 79. An α had initial kinetic energy of about 6 MeV. When an α was closest to a gold nucleus, it stopped completely by the electric repulsive force between the two nuclei. Assume all the α kinetic energy was converted to electric potential energy. The closest distance (center to center) between the two nuclei is _____ x 10-15 meter.
Chapter 29 Solutions
College Physics: A Strategic Approach (3rd Edition)
Ch. 29 - Prob. 1CQCh. 29 - Prob. 2CQCh. 29 - Prob. 3CQCh. 29 - Prob. 4CQCh. 29 - Prob. 5CQCh. 29 - Prob. 6CQCh. 29 - Prob. 7CQCh. 29 - Prob. 8CQCh. 29 - Prob. 9CQCh. 29 - Prob. 10CQ
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- (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_forwardSuppose the alpha particle in a Rutherford scattering experiment is replaced with a proton of the same initial kinetic energy and also headed directly toward the nucleus of the gold atom. (a) Will the distance from the center of the nucleus at which the proton stops be greater than, less than, or the same as that of the alpha particle? (b) If, instead, we switch the target to a nucleus with a larger value of Z, is the stopping distance of the alpha particle greater than, less than, or the same as with the gold target?arrow_forwardSuppose 8.5x10−19 J of work is done to bring the nucleus of an atom from some infinite distance away to a particular location near a fixed positive charge. What would the electric potential energy of the nucleus be? The nucleus is released an returns to the very far original position. Considering the kinetic energy of the nucleus and a mass of 6.6x10−27 kg, which is closest to the velocity of the nucleus? Make sure to answer both parts of the question and show all work.arrow_forward
- nucleus emits an alpha particle with kinetic evergy = 4.20 MeV. What is the daughter nucleus, and what is the approximate atomic mass (in u) of the daughter atom? ignore recoil of the daughter nucleus. if the image doesnt come up properly it is 92^238 Uarrow_forwardStarting from the Clasius-Clapeyron equation, Lo + constant kBT log p and assuming the latent heat of evaporation per gram of water to be constant, and equal to L= 2260 J/g , find the boiling temperature of water in degrees Celcius at a pressure of 50.0 atm. Each oxygen atom has 8 protons and 8 neutrons in its nucleus. The boiling temperature of water at a pressure of 50.0 atm is around 250 °C. How do you account for the difference? R = 8.31 //(mol · K) , m,-1.67x1027 kg , NA=6.02x1023 mol-1arrow_forwardPhotoarrow_forward
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