Answer
$\lambda_n=\frac{nh^2}{2\pi mKe^2}$
Work Step by Step
We know that
$\lambda_n=\frac{h}{P_n}$
$\lambda_n=\frac{h}{mv_n}$
$\implies \lambda_n=\frac{h}{m(2\pi Ke^2/nh)}$
This can be simplified as
$\lambda_n=\frac{nh^2}{2\pi mKe^2}$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 31 - Atomic Physics - Problems and Conceptual Exercises - Page 1113: 36
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