College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1042: 17

Answer

$\Delta V = 4970~V$

Work Step by Step

We can find the minimum potential difference $\Delta V$ applied to the tube: $q~\Delta V = \frac{hc}{\lambda}$ $\Delta V = \frac{hc}{q~\lambda}$ $\Delta V = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.6\times 10^{-19}~C)(0.250\times 10^{-9}~m)}$ $\Delta V = 4970~V$
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