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 1402: 41.6

Answer

6.16 MeV.

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 $\frac{(n_x^2+ n_y^2+n_z^2)\pi^2\hbar^2}{2mL^2}$ 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 $\Delta E = 6.16 MeV$.
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