Answer
$5$
Work Step by Step
For given principal quantum number $n$, the allowed values of orbital quantum number $l$ are $l=0,1,2,...,(n-1)$
An electron in a multielectron atom has $l=4$
Therefore, the smallest possible value of $n$ is $5$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 40 - All About Atoms - Problems - Page 1247: 7b
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