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 1115: 40


(a) $E_0 = 2.3~eV$ (b) $\lambda_1 = 244~nm$

Work Step by Step

(a) We can write an expression for the maximum kinetic energy: $K_{max} = \frac{hc}{\lambda} - E_0,~~~$ (where $E_0$ is the work function) Let $E_1 = \frac{hc}{\lambda_1}$ be the initial energy of each photon of light. Then: $K_{max} = E_1 - E_0$ If the wavelength is increased by 50%, then the energy of each photon decreases to $\frac{2~E_1}{3}$ We can set up two equations as follows: (1) $K_{max} = E_1 - E_0 = 2.8~eV$ (1) $E_1 - E_0 = 2.8~eV$ (2) $K_{max} = \frac{2~E_1}{3} - E_0 = 1.1~eV$ (2) $-E_1 + 1.5~E_0 = -1.65~eV$ We can add the two equations to find $E_0$: $0.5~E_0 = 1.15~eV$ $E_0 = 2.3~eV$ (b) We can find the initial wavelength: $K_{max} = \frac{hc}{\lambda_1} - E_0 = 2.8~eV$ $\frac{hc}{\lambda_1} = 2.8~eV+E_0$ $\lambda_1 = \frac{hc}{2.8~eV+E_0}$ $\lambda_1 = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(2.8~eV+2.3~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda_1 = 2.44\times 10^{-7}~m$ $\lambda_1 = 244~nm$
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