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: 6.30

Answer

The longest wavelength of radiation that possesses the necessary energy to break the bond is $494.3nm$. A radiation with that wavelength is a form of visible light.

Work Step by Step

*Strategy: Since we already know the minimum energy in $kJ/mol$ required to break the bond ($E_{min}$), we can find the minimum energy required to break a single bond ($E_{b, min}$). From there, we could deduce the longest wavelength of radiation that possesses the necessary energy to break the bond ($\lambda_{max}$). - Step 1: Find the minimum energy required to break a single bond $E_{b, min}$ We have $E_{min}=242kJ/mol=2.42\times10^5J/mol$ and the number of bonds considered here: $N=1 mol Cl_2\approx6.022\times10^{23}$ bonds. $E_{b, min}=\frac{E_{min}}{N}=\frac{2.42\times10^5}{6.022\times10^{23}}\approx4.019\times10^{-19}J/bond$ - Step 2: Find the longest wavelength of radiation having the necessary energy to break the bond $\lambda_{max}$ We would consider this formula $$E_{b}=h\times\nu=\frac{h\times c}{\lambda}$$ $E_{b}$ : the energy required to break a single bond $h$ : Planck's constant ($h\approx6.626\times10^{-34}J.s$) $\nu$ : frequency of radiation $c$ : speed of light in a vacuum ($c\approx2.998\times10^8m/s$) $\lambda$ : wavelength of radiation Because both $h$ and $c$ are constants, and $E_{b}$ and $\lambda$ are inversely proportional, the minimum value of $E_b$ would require the maximum value of $\lambda$ (to break the bond). Therefore: $$E_{b, min}=\frac{h\times c}{\lambda_{max}}$$ In other words, $\lambda_{max}=\frac{h\times c}{E_{b, min}}= \frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.019\times10^{-19}}\approx4.943\times10^{-7}m\approx494.3nm$ A radiation with a wavelength of $494.3nm$ lies in the wavelength range of visible light. Therefore, this radiation is a form of visible light.
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