Answer
$2.86eV$
Work Step by Step
The required energy can be determined as
$\Delta E=E_5-E_2$
$\implies \Delta E=(-13.6eV)(\frac{1}{5^2}-\frac{1}{2^2})$
$\Delta E=2.86eV$
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 1112: 18
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