University Physics with Modern Physics (14th Edition)

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

Chapter 41 - Quantum Mechanics II: Atomic Structure - Problems - Exercises - Page 1401: 41.2


See explanation.

Work Step by Step

For a particle in a cubical box of side length L, with quantum numbers $(n_x, n_y, n_z)$, the energy is $\frac{(n_x^2+ n_y^2+n_z^2)\pi^2\hbar^2}{2mL^2}$. Here, we are told that $L^3=\frac{4\pi a_o^3}{3}$, where $a_o$ is the Bohr radius. This means that $L=8.53\times10^{-11}m$. Therefore $E_{1,1,1}= \frac{3\pi^2\hbar^2}{2mL^2}$, We also see that $E_{2,1,1}= \frac{6\pi^2\hbar^2}{2mL^2}$, $\Delta E= \frac{3\pi^2\hbar^2}{2mL^2}$ Evaluating this expression, we find $\Delta E = 155 eV$. In contrast, in the Bohr model, $E_n=-\frac{13.6eV}{n^2}$ so the energy difference between the first and second levels is $\Delta E =-13.6eV (\frac{1}{2^2}-\frac{1}{1^2})=10.2 eV$. We conclude that a particle in a box is not a great model for a hydrogen atom.
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