## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 27 - Problems - Page 1045: 87

#### Answer

The longest three wavelengths in the Lyman series are $121.5~nm$, $102.6~nm$, and $97.23~nm$

#### Work Step by Step

The longest three wavelengths in the Lyman series would be the three transitions with the least amount of energy, that is, the transitions from level 2 to the ground state, from level 3 to the ground state, and from level 4 to the ground state. We can find the wavelength of the photon emitted by a transition from level 2 to the ground state: $\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ $\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{2^2})$ $\frac{1}{\lambda} = R~(\frac{3}{4})$ $\lambda = \frac{4}{3~R}$ $\lambda = \frac{4}{(3)~(1.097\times 10^7~m^{-1})}$ $\lambda = 1.215\times 10^{-7}~m$ $\lambda = 121.5~nm$ We can find the wavelength of the photon emitted by a transition from level 3 to the ground state: $\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ $\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{3^2})$ $\frac{1}{\lambda} = R~(\frac{8}{9})$ $\lambda = \frac{9}{8~R}$ $\lambda = \frac{9}{(8)~(1.097\times 10^7~m^{-1})}$ $\lambda = 1.026\times 10^{-7}~m$ $\lambda = 102.6~nm$ We can find the wavelength of the photon emitted by a transition from level 4 to the ground state: $\frac{1}{\lambda} = R~(\frac{1}{n_f^2}-\frac{1}{n_i^2})$ $\frac{1}{\lambda} = R~(\frac{1}{1^2}-\frac{1}{4^2})$ $\frac{1}{\lambda} = R~(\frac{15}{16})$ $\lambda = \frac{16}{15~R}$ $\lambda = \frac{16}{(15)~(1.097\times 10^7~m^{-1})}$ $\lambda = 0.9723\times 10^{-7}~m$ $\lambda = 97.23~nm$ The longest three wavelengths in the Lyman series are $121.5~nm$, $102.6~nm$, and $97.23~nm$.

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