# Chapter 5 - Electron Configurations and the Periodic Table - Questions for Review and Thought - Topical Questions: 16

$E = 3.03*10^{-19}J$

#### Work Step by Step

We know that: $E = \frac{hc}{ \lambda}=h \nu$ $E = 6.625*10^{-34}*4.57*10^{14}$ $E = 3.03*10^{-19}J$

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