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 - Problems - Page 799: 23

Answer

$KE_{max}=0.92eV$ $v=5.7\times10^5m/s$

Work Step by Step

In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$. Use equation 27–5b to calculate the maximum kinetic energy. $$KE_{max}=hf-W_o=\frac{hc}{\lambda}-W_o$$ $$KE_{max}=\frac{1240\;eV \cdot nm }{365nm}-2.48eV=0.92eV$$ To calculate the electron speed, set this energy equal to $\frac{1}{2}mv^2$ and solve for the speed v of $5.7\times10^5m/s$. The energy is much less than the electron's rest energy, so the classical expression for KE is sufficient.
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