Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 38 - Quantization - Exercises and Problems - Page 1154: 53

Answer

$L = 18~fm$

Work Step by Step

We can write a general equation for the energy: $E_n = \frac{n^2~h^2}{8mL^2}$ The energy difference between the $n = 2$ state and the $n = 1$ state is: $\Delta E = \frac{3~h^2}{8mL^2} = 2.0~MeV$ We can find the length of the box: $\frac{3~h^2}{8mL^2} = 2.0~MeV$ $L^2 = \frac{3~h^2}{(8m)(2.0~MeV)}$ $L = \sqrt{\frac{3~h^2}{(8m)(2.0~MeV)}}$ $L = \sqrt{\frac{(3)~(6.626\times 10^{-34}~J~s)^2}{(8)(1.67\times 10^{-27}~kg)(2.0\times 10^6~eV)(1.6\times 10^{-19}~J/eV)}}$ $L = 1.8\times 10^{-14}~m$ $L = 18~fm$
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