College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1041: 11


$\phi = 7.3\times 10^{-19}~J$

Work Step by Step

We can find the magnitude of the energy removed from the electron by the stopping potential: $E = q~\Delta V$ $E = (1.6\times 10^{-19}~C)(1.1~V)$ $E = 1.76\times 10^{-19}~J$ This energy is equal to the maximum kinetic energy of the electron as it is ejected from the tungsten surface. We can find the work function $\phi$: $K_{max} = \frac{hc}{\lambda}-\phi$ $\phi = \frac{hc}{\lambda}-K_{max}$ $\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~/s)}{220\times 10^{-9}~m}-1.76\times 10^{-19}~J$ $\phi = 9.035\times 10^{-19}~J-1.76\times 10^{-19}~J$ $\phi = 7.3\times 10^{-19}~J$
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