Answer
326 nm.
Work Step by Step
The photon’s minimum energy is the 3.80 eV bond strength.
$$E=hf=h\frac{c}{\lambda}$$
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{3.80eV}=3.26\times10^{-7}m$$
University Physics with Modern Physics (14th Edition)
by Young, Hugh D.; Freedman, Roger A.
Published by
Pearson
ISBN 10:
0321973615
ISBN 13:
978-0-32197-361-0
Chapter 39 - Particles Behaving as Waves - Problems - Exercises - Page 1314: 39.18
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