University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
14th Edition
ISBN: 9780134265414
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
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Chapter 40, Problem 40.50P
To determine
To show: That the impulse exerted by the wall at
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T5
PROBLEM 2: Consider neutrons slowing down by elastic scattering from 1.0 MeV to 0.025 eV
in large spatially homogeneous regions.
(a) Estimate the number of scattering events required for these neutrons to slow down in large
regions of hydrogen (A = 1), iron (A = 56), and uranium (A = 238).
(b) Suppose that in each neutron-nucleus collision, the probability that a neutron is captured
is 0.001 (independent of energy). Estimate the probability that these neutrons in regions
238) will not be captured while
of hydrogen (A
1), iron (A
56), and uranium (A
slowing down from 1.0 MeV to 0.025 e V.
Note: In practice, the probability that a neutron with energy E will be captured depends on E.
This makes realistic calculations of the non-capture probability considerably more difficult.
Electron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino):
e − + 7 B e + ⟶ 7 L i + ν
The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably:
7 B e ⟶ 7 L i + ν
The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.)
The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (all the energy released goes into the lighter particle. c = 300,000 km/s)
Chapter 40 Solutions
University Physics with Modern Physics, Volume 2 (Chs. 21-37); Mastering Physics with Pearson eText -- ValuePack Access Card (14th Edition)
Ch. 40.1 - Does a wave packet given by Eq. (40.19) represent...Ch. 40.2 - Prob. 40.2TYUCh. 40.3 - Prob. 40.3TYUCh. 40.4 - Prob. 40.4TYUCh. 40.5 - Prob. 40.5TYUCh. 40.6 - Prob. 40.6TYUCh. 40 - Prob. 40.1DQCh. 40 - Prob. 40.2DQCh. 40 - Prob. 40.3DQCh. 40 - Prob. 40.4DQ
Ch. 40 - If a panicle is in a stationary state, does that...Ch. 40 - Prob. 40.6DQCh. 40 - Prob. 40.7DQCh. 40 - Prob. 40.8DQCh. 40 - Prob. 40.9DQCh. 40 - Prob. 40.10DQCh. 40 - Prob. 40.11DQCh. 40 - Prob. 40.12DQCh. 40 - Prob. 40.13DQCh. 40 - Prob. 40.14DQCh. 40 - Prob. 40.15DQCh. 40 - Prob. 40.16DQCh. 40 - Prob. 40.17DQCh. 40 - Prob. 40.18DQCh. 40 - Prob. 40.19DQCh. 40 - Prob. 40.20DQCh. 40 - Prob. 40.21DQCh. 40 - Prob. 40.22DQCh. 40 - Prob. 40.23DQCh. 40 - Prob. 40.24DQCh. 40 - Prob. 40.25DQCh. 40 - Prob. 40.26DQCh. 40 - Prob. 40.27DQCh. 40 - Prob. 40.1ECh. 40 - Prob. 40.2ECh. 40 - Prob. 40.3ECh. 40 - Prob. 40.4ECh. 40 - Prob. 40.5ECh. 40 - Prob. 40.6ECh. 40 - Prob. 40.7ECh. 40 - Prob. 40.8ECh. 40 - Prob. 40.9ECh. 40 - Prob. 40.10ECh. 40 - Prob. 40.11ECh. 40 - Prob. 40.12ECh. 40 - Prob. 40.13ECh. 40 - Prob. 40.14ECh. 40 - Prob. 40.15ECh. 40 - Prob. 40.16ECh. 40 - Prob. 40.17ECh. 40 - Prob. 40.18ECh. 40 - Prob. 40.19ECh. 40 - Prob. 40.20ECh. 40 - Prob. 40.21ECh. 40 - Prob. 40.22ECh. 40 - Prob. 40.23ECh. 40 - Prob. 40.24ECh. 40 - Prob. 40.25ECh. 40 - Prob. 40.26ECh. 40 - Prob. 40.27ECh. 40 - Prob. 40.28ECh. 40 - Prob. 40.29ECh. 40 - Prob. 40.30ECh. 40 - Prob. 40.31ECh. 40 - Prob. 40.32ECh. 40 - Prob. 40.33ECh. 40 - Prob. 40.34ECh. 40 - Prob. 40.35ECh. 40 - Prob. 40.36ECh. 40 - Prob. 40.37ECh. 40 - Prob. 40.38ECh. 40 - Prob. 40.39ECh. 40 - Prob. 40.40ECh. 40 - Prob. 40.41ECh. 40 - Prob. 40.42PCh. 40 - Prob. 40.43PCh. 40 - Prob. 40.44PCh. 40 - Prob. 40.45PCh. 40 - Prob. 40.46PCh. 40 - Prob. 40.47PCh. 40 - Prob. 40.48PCh. 40 - Prob. 40.49PCh. 40 - Prob. 40.50PCh. 40 - Prob. 40.51PCh. 40 - Prob. 40.52PCh. 40 - Prob. 40.53PCh. 40 - Prob. 40.54PCh. 40 - Prob. 40.55PCh. 40 - Prob. 40.56PCh. 40 - Prob. 40.57PCh. 40 - Prob. 40.58PCh. 40 - Prob. 40.59PCh. 40 - Prob. 40.60PCh. 40 - Prob. 40.61PCh. 40 - Prob. 40.62PCh. 40 - Prob. 40.63PCh. 40 - Prob. 40.64CPCh. 40 - Prob. 40.65CPCh. 40 - Prob. 40.66CPCh. 40 - Prob. 40.67PPCh. 40 - Prob. 40.68PPCh. 40 - Prob. 40.69PPCh. 40 - Prob. 40.70PP
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- Electron capture is a variant on beta-radiation. The lightest nucleus to decay by electron capture is 7Be -- beryllium-7. The daughter nucleus is 7Li -- lithium-7. The electron is transformed into a massless particle (a neutrino): e − + 7 B e + ⟶ 7 L i + ν The initial electron is bound in the atom, so the beryllium mass includes the electron. In fact, since the electron starts bound in the atom, a more-accurate statement of the nuclear reaction is probably: 7 B e ⟶ 7 L i + ν The masses are beryllium: 7.016929 u, and lithium: 7.016003 u, and refer to the neutral atom as a whole. (Use uc and uc2 as your momentum and energy units -- but carry them along in your calculation.) The initial beryllium atom is stationary. Calculate the speed of the final lithium nucleus in km/s. (You will make life much easier for yourself if you recognize that practically all the energy released goes into the lighter particle. c = 300,000 km/s)arrow_forwardAn atom of mass m₁ = m moves in the positive x direc- tion with speed v₁ = v. It collides with and sticks to an atom of mass m₂ = 2m moving in the positive y direction with speed v₂ = 2v/3. Find the resultant speed and direc- tion of motion of the combination, and find the kinetic energy lost in this inelastic collision.arrow_forwardFirst part is already answered but the process would still be appreciated.arrow_forward
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