Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 38 - Quantization - Exercises and Problems - Page 1114: 5

Answer

$E_0 = 1.77~eV$

Work Step by Step

When $\lambda = 700~nm$, we can assume that $K_{max} = 0$ We can find the maximum work function of the cathode: $K_{max} = \frac{h~c}{\lambda} - E_0$ $E_0 = \frac{h~c}{\lambda}$ $E_0 = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{700\times 10^{-9}~m}$ $E_0 = 2.84\times 10^{-19}~J$ $E_0 = (2.84\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E_0 = 1.77~eV$
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