Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 6 - Electronic Structure of Atoms - Exercises - Page 250: 6.34a

Answer

The minimum frequency of light necessary to emit electrons from titanium is $1.047\times10^{15}s^{-1}$.

Work Step by Step

According to the formula: $$E=h\times \nu$$ Since $h$ is a constant, the minimum value of $E$ ($E_{min}$) would be achieved at the minimum value of $\nu$ ($\nu_{min}$). In other words, $$E_{min}=h\times\nu_{min}$$ From the question, we have: - Minimum energy of photon for the metal to emit electrons: $E_{min}=6.94\times10^{-19}J$ - Also, the Planck's constant: $h\approx6.626\times10^{-34}J.s$ Therefore, the minimum frequency of light necessary to emit electrons from titanium is: $\nu_{min}=\frac{E_{min}}{h}=\frac{6.94\times10^{-19}}{6.626\times10^{-34}}\approx1.047\times10^{15}s^{-1}$
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