College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 27 - Problems - Page 1042: 18

Answer

$f = 1.1\times 10^{19}~Hz$

Work Step by Step

We can find the cut off frequency: $hf = q~\Delta V$ $f = \frac{q~\Delta V}{h}$ $f = \frac{(1.6\times 10^{-19}~C)(46000~V)}{6.626\times 10^{-34}~J~s}$ $f = 1.1\times 10^{19}~Hz$
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