Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Question
Chapter 31, Problem 16P
To determine
The reasons for which one reaction will occur and other reaction will not occur.
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Students have asked these similar questions
1) For each of the following reactions work out the fastest interaction through which
the conservation laws allow it to proceed. Explain your answers. If the reaction is
forbidden by all interactions explain why:
-+
a. pn++μ+ +μ¯
b. Aºp+e¯
c. μ΄ →e + V
d. p+p→y+Y
e. Kºn++ π + π° +7°
f. π+pAº + Kº
g. Aºn+p
h. pnº+e+ + ve
i. n→p+e+V₂
The first of the following two reactions can occur, but thesecond cannot. Explain.KS0 → π+ + π- (can occur)L0 → π+ + π- (cannot occur)
Considering the conservation laws, why are the following reactions forbidden?
i.
Vụ + p → H +n
A> n* + e + ve
K*→ T° + µ + vụ
ii.
iii.
iv.
e +p >p + e*
V.
Chapter 31 Solutions
Principles of Physics: A Calculus-Based Text
Ch. 31.2 - Given the identification of the particles in...Ch. 31.5 - Prob. 31.2QQCh. 31.5 - Prob. 31.3QQCh. 31.5 - Prob. 31.4QQCh. 31.9 - Prob. 31.5QQCh. 31 - Prob. 1OQCh. 31 - Prob. 2OQCh. 31 - Prob. 3OQCh. 31 - Prob. 4OQCh. 31 - Prob. 5OQ
Ch. 31 - Prob. 6OQCh. 31 - Prob. 7OQCh. 31 - Prob. 8OQCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQCh. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 10CQCh. 31 - Prob. 11CQCh. 31 - Prob. 12CQCh. 31 - Prob. 13CQCh. 31 - Prob. 1PCh. 31 - Prob. 2PCh. 31 - Prob. 3PCh. 31 - Prob. 4PCh. 31 - Prob. 5PCh. 31 - Prob. 6PCh. 31 - Prob. 7PCh. 31 - Prob. 8PCh. 31 - Prob. 9PCh. 31 - Prob. 10PCh. 31 - Prob. 11PCh. 31 - Prob. 12PCh. 31 - Prob. 13PCh. 31 - Prob. 14PCh. 31 - Prob. 15PCh. 31 - Prob. 16PCh. 31 - Prob. 17PCh. 31 - Prob. 18PCh. 31 - Prob. 19PCh. 31 - Prob. 20PCh. 31 - Prob. 21PCh. 31 - Prob. 22PCh. 31 - Prob. 23PCh. 31 - Prob. 24PCh. 31 - Prob. 25PCh. 31 - Prob. 26PCh. 31 - Prob. 27PCh. 31 - Prob. 28PCh. 31 - Prob. 29PCh. 31 - Prob. 30PCh. 31 - Prob. 31PCh. 31 - Prob. 32PCh. 31 - Prob. 33PCh. 31 - Prob. 34PCh. 31 - Prob. 35PCh. 31 - Prob. 36PCh. 31 - Prob. 37PCh. 31 - Prob. 38PCh. 31 - Prob. 39PCh. 31 - Prob. 40PCh. 31 - Prob. 41PCh. 31 - Prob. 42PCh. 31 - Prob. 43PCh. 31 - Prob. 44PCh. 31 - Prob. 45PCh. 31 - Prob. 46PCh. 31 - Prob. 47PCh. 31 - Prob. 48PCh. 31 - Prob. 49PCh. 31 - Prob. 50PCh. 31 - Prob. 51PCh. 31 - Prob. 52PCh. 31 - Prob. 53PCh. 31 - Prob. 54PCh. 31 - Prob. 55PCh. 31 - Prob. 56PCh. 31 - Prob. 57PCh. 31 - Prob. 58PCh. 31 - Prob. 59PCh. 31 - Prob. 60PCh. 31 - Prob. 61PCh. 31 - Prob. 62PCh. 31 - Prob. 63PCh. 31 - Prob. 64PCh. 31 - Prob. 65P
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- Integrated Concepts: (a) What temperature gas would have atoms moving fast enough to bring two 3He nuclei into contact? Note that, because both are moving, the average kinetic energy only needs to be half the electric potential energy of these doubly charged nuclei when just in contact with one another. (b) Does this high temperature imply practical difficulties for doing this in controlled fusion?arrow_forwardUnreasonable Results A particle physicist discovers a neutral particle with a mass at 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?arrow_forward(a) Calculate the energy released in the a decay of 238U. (b) What fraction of the mass at a single 238U is destroyed in the decay? The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is laws for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?arrow_forward
- Determine which of the following reactions can occur. For those that cannot occur, determine the conservation law (or laws) violated. (a) p→ ㅠ + mo (b) p + p → p + p + 7° (c) p+p - p+ n* (e) n → p+e- + ī, (d) * → ut + v. (f) → ut + narrow_forwardWhich of the following reactions cannot occur? (Select all that apply.) 1)π- → μ- + vμ 2)μ+ → e+ + ve + vμ 3)p + p → p + p + p 4)n → p + e- + vearrow_forwardFor the following two reactions, the first may occur but the second cannot. Explain. K0 → π++π− (can occur) ^0 → π+ + π− (cannot occur)arrow_forward
- Explain why each of the following reactions is forbidden. (a) p + p → p + p + n (b) p + p → p + π+ + γarrow_forwardFor the following two reactions, the first may occur but the second cannot. Explain.arrow_forwardConsider the reactions at the same energy + π' + p → Σ+ + K+ π + p→Σ + K+ π+P→ E + Kº Assuming that the isospin amplitude a₁/2 < a3/2, show that the cross sections for the reactions will be in the ratio 9:1:2arrow_forward
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