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: 42

Answer

(a) The work function is $~~4.14~eV$ (b) $h = 4.0\times 10^{-15}~eV~s$

Work Step by Step

(a) From the graph, we can see that the threshold frequency is $1.0\times 10^{15}~Hz$ Thus, the energy of each photon of this frequency is equal to the work function. We can find the energy of each photon: $E = h~f$ $E = (6.626\times 10^{-34}~J~s)(1.0\times 10^{15}~Hz)$ $E = 6.626\times 10^{-19}~J$ $E = (6.626\times 10^{-19}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$ $E = 4.14~eV$ The work function is $~~4.14~eV$ (b) We can write an equation for the stopping potential: $K_{max} = hf- E_0$ $\Delta V~e = hf- E_0$ $\Delta V = \frac{h}{e}~f- \frac{E_0}{e}$ This equation is a linear equation where $\frac{h}{e}$ is the slope of the $\Delta V$ versus $f$ graph. We can find $h$: $\frac{h}{e} = \frac{8.0~V - 0~V}{(3.0\times 10^{15}~Hz)-(1.0\times 10^{15}~Hz)}$ $h = (e)~(4.0\times 10^{-15}~V~s)$ $h = 4.0\times 10^{-15}~eV~s$
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