Answer
Nickel.
Work Step by Step
Apply equation 41.48, relating the x-ray energy to Z.
$$E_{K_\alpha}=(Z-1)^2(10.2\;eV)$$
$$7460eV=(Z-1)^2(10.2\;eV)$$
$$Z=1+\sqrt{\frac{7460\;eV }{10.2\;eV}}\approx 28$$
The element is nickel.
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 41 - Quantum Mechanics II: Atomic Structure - Problems - Exercises - Page 1403: 41.36
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
