College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1041: 6


(a) $\phi = 4.31~eV$ (b) $K_{max} = 4.56~eV$

Work Step by Step

(a) We can find the work function $\phi$: $\phi = \frac{hc}{\lambda_0}$ $\phi = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{288\times 10^{-9}~m}$ $\phi = 6.9\times 10^{-19}~J$ $\phi = (6.9\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $\phi = 4.31~eV$ (b) We can find the maximum kinetic energy: $K_{max} = \frac{hc}{\lambda}-\phi$ $K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{140\times 10^{-9}~m}-4.31~eV$ $K_{max} = (1.42\times 10^{-18}~J)-(4.31~eV)$ $K_{max} = (1.42\times 10^{-18}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})-(4.31~eV)$ $K_{max} = 4.56~eV$
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