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: 6.41b

Answer

- When $n_i=3, \lambda=656nm$. This is the red line near the right end of the spectra line. - When $n_i=4, \lambda=486nm$. This is the blue-green line near the $500nm$ point. - When $n_i=5, \lambda=433.9nm$. This is the blue-violet line near the $450nm$ point.

Work Step by Step

*When $n_f=2$, we have: $\Delta E= E_f - E_i=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{n^2_i})$ 1) $n_i=3$: $\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{9})\approx-3.028\times10^{-19}J$ So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is: $E_{photon}=|\Delta E|\approx3.028\times10^{-19}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.028\times10^{-19}}\approx6.56\times10^{-7}m\approx656nm$ This is the red line near the right end of the line spectra. 2) $n_i=4$: $\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{16})=-4.0875\times10^{-19}J$ So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is: $E_{photon}=|\Delta E|=4.0875\times10^{-19}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.0875\times10^{-19}}\approx4.86\times10^{-7}m\approx486nm$ This is the blue-green line near the point $500nm$. 3) $n_i=5$: $\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{25})=-4.578\times10^{-19}J$ So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is: $E_{photon}=|\Delta E|=4.578\times10^{-19}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.578\times10^{-19}}\approx4.339\times10^{-7}m\approx433.9nm$ This is the blue-violet line near the $450nm$ point.
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