Chemistry: The Central Science (13th Edition)

Published by Prentice Hall

Chapter 6 - Electronic Structure of Atoms - Exercises: 6.42b

Answer

For $n_i=2, \lambda=121.5nm$ For $n_i=3, \lambda=102.5nm$ For $n_i=4, \lambda=97.2nm$

Work Step by Step

*When $n_f=1$, we have: $\Delta E= E_f - E_i=(-2.18\times10^{-18})\times(\frac{1}{1^2}-\frac{1}{n^2_i})=(-2.18\times10^{-18})\times(1-\frac{1}{n^2_i})$ 1) $n_i=2$: $\Delta E=(-2.18\times10^{-18})\times(1-\frac{1}{4})=-1.635\times10^{-18}J$ So, the energy of a photon of light emitted when electron moves from n = 2 to n = 1 is: $E_{photon}=|\Delta E|=1.635\times10^{-18}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{1.635\times10^{-18}}\approx1.215\times10^{-7}m\approx121.5nm$ 2) $n_i=3$: $\Delta E=(-2.18\times10^{-18})\times(1-\frac{1}{9})=-1.938\times10^{-18}J$ So, the energy of a photon of light emitted when electron moves from n = 3 to n = 1 is: $E_{photon}=|\Delta E|=1.938\times10^{-18}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{1.938\times10^{-18}}\approx1.025\times10^{-7}m\approx102.5nm$ 3) $n_i=4$: $\Delta E=(-2.18\times10^{-18})\times(1-\frac{1}{16})\approx-2.044\times10^{-18}J$ So, the energy of a photon of light emitted when electron moves from n = 4 to n = 1 is: $E_{photon}=|\Delta E|=2.044\times10^{-18}J$ The wavelength of light emitted is: $\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{2.044\times10^{-18}}\approx9.719\times10^{-8}m\approx97.2nm$

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