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: 51

Answer

$L = 0.354~nm$

Work Step by Step

We can write a general equation for the energy: $E_n = \frac{n^2~h^2}{8mL^2}$ Note the following relationships: $\sqrt{\frac{27~eV}{12~eV}} = \frac{3}{2}$ $\sqrt{\frac{48~eV}{12~eV}} = \frac{4}{2}$ Thus the $12~eV$ energy level is the $n = 2$ state. We can find the length of the box: $E_n = \frac{n^2~h^2}{8mL^2}$ $E_2 = \frac{(2)^2~h^2}{8mL^2}$ $L^2 = \frac{(2)^2~h^2}{8m~E_2}$ $L = \frac{(2)~h}{\sqrt{8m~E_2}}$ $L = \frac{(2)~(6.626\times 10^{-34}~J~s)}{\sqrt{(8)(9.109\times 10^{-31}~kg)~(12~eV)(1.6\times 10^{-19}~J/eV)}}$ $L = 0.354\times 10^{-9}~m$ $L = 0.354~nm$
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