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 1403: 41.27

Answer

See explanation. The ten lowest energy levels for electrons in neon have n = 1 or n = 2.

Work Step by Step

For n = 1, $l$ can only be 0, and $m_l$ can only be 0. Here are the possible $(n, l, m_l,m_s)$ pairs for n =1: $$(1,0,0,+\frac{1}{2}) (1,0,0,-\frac{1}{2})$$ For n = 2, $l$ can take on the values 1 and 0. For $l=1$, $m_l$ can be $\pm l$. Here are the possible $(n, l, m_l,m_s)$ pairs for n =1: $$(2,1,0,+\frac{1}{2}) (2,1,0,-\frac{1}{2})$$ $$(2,1,1,+\frac{1}{2}) (2,1,1,-\frac{1}{2})$$ $$(2,1,-1,+\frac{1}{2}) (2,1,-1,-\frac{1}{2})$$ $$(2,0,0,+\frac{1}{2}) (2,0,0,-\frac{1}{2})$$ Those cover the 10 electrons in neon.
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