Answer
Please see the work below.
Work Step by Step
We know that
$E_1=-(2.18\times 10^{-18}J(2)^2)(6.02\times 10^{23})=-5.25\times 10^{3}\frac{KJ}{mol}$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 8 - Periodic Relationships Among the Elements - Questions & Problems - Page 362: 8.57
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