College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1041: 5


(a) $K_{max} = 0.848~eV$ (b) $\lambda = 575~nm$

Work Step by Step

(a) We can find the maximum kinetic energy: $K_{max} = hf-\phi$ $K_{max} = \frac{hc}{\lambda}-\phi$ $K_{max} = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{413\times 10^{-9}~m}-2.16~eV$ $K_{max} = (4.813\times 10^{-19}~J)-(2.16~eV)$ $K_{max} = (4.813\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})-(2.16~eV)$ $K_{max} = (3.008~eV)-(2.16~eV)$ $K_{max} = 0.848~eV$ (b) We can find the threshold wavelength: $E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.16~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda = 5.75\times 10^{-7}~m$ $\lambda = 575~nm$
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