University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 43 - Nuclear Physics - Problems - Exercises - Page 1475: 43.2

Answer

(a) $\Delta E = 2.77 \times 10^{-7} eV$. Anti-parallel configuration has lower energy. (b) $ f = 6.689 \times 10^7 Hz $ $\lambda = 4.48 m$

Work Step by Step

$(a)$ We use the equation $\Delta E = 2 \mu_zB$, to find the transition energy between states. where $\mu_z = 1.913 \mu_n$ t and $\mu_n = 3.1525 \times 10^{8} eV/T$ $\Delta E = 2 \mu_zB$ $\Delta E = 2 (1.913)(3.1525 \times 10^{-8} eV/T)(2.30T)$ $\Delta E = 2.77 \times 10^{-7} eV$ *The antiparallel configuration is lower energy in neutron because the $\overrightarrow μ $ and $\overrightarrow S$ are in opposite directions. This result is smaller than configuration in proton. $(b)$ Frequency of photon $ f = \frac{\Delta E}{h}$ $ f = \frac{(2.77 \times 10^{-7} eV)(1.6 \times 10^{-19} J/eV)}{6.626 \times10^{-34} J.s}$ $ f = 6.689 \times 10^7 Hz $ or $ 66.89 MHz$ Wavelength of photon $\lambda = \frac{c}{f}$ $\lambda = \frac{3.0 \times 10^8 m/s }{ 6.689 \times 10^7 Hz }$ $\lambda = 4.48 m$
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