Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 27 - Early Quantum Theory and Models of the Atom - General Problems - Page 801: 81

Answer

0.33 volts.

Work Step by Step

The stopping potential $V_0$ is the voltage such that $eV_0$ equals the maximum kinetic energy. Apply equation 27–5b to first find the work function. $$KE_{max}=eV_0=\frac{hc}{\lambda_0}-W_0$$ $$ W_0= \frac{hc}{\lambda_0}- eV_0 $$ Now find the stopping potential for the higher wavelength light. $$ eV_1=\frac{hc}{\lambda_1}-W_0= hc (\frac{1}{\lambda_1}-\frac{1}{\lambda_0})+eV_0$$ $$ eV_1= (1240 eV \cdot nm) (\frac{1}{440nm}-\frac{1}{270nm })+2.10eV=0.33eV$$ The new stopping potential is 0.33 volts.
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