General Chemistry (4th Edition)

Published by University Science Books
ISBN 10: 1891389602
ISBN 13: 978-1-89138-960-3

Chapter 4 Early Quantum Theory - Problems: 30

Answer

$5.97\cdot 10^{14}s^{-1}$

Work Step by Step

We have \begin{align*}\frac{m_ev^2}{2}&=E_k=h\left(\frac{c}{39\cdot 10^{-8}m}-\nu_0\right)\\\Longrightarrow \nu_0&=\frac{3\cdot 10^8 ms^{-1}}{39\cdot 10^{-8}m}-\frac{9.1\cdot 10^{-31}kg\cdot 25\cdot 10^{10}m^2s^{-2}}{2\cdot 6.63\cdot 10^{-34}m^2kgs^{-1}}\\& =7.69\cdot 10^{14}Hz-1.715\cdot 10^{15}Hz=5.97\cdot 10^{14}s^{-1}\end{align*}
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