Quantum Physics

28.3 Th Bohr Theory of Hydrogen 909We can use this expression to evaluate the wavelengths for the various series in thehydrogen spectrum. For example, in the Balmer series, n f 2 and n i 3, 4, 5, . . .(Eq. 28.1). For the Lyman series, we take n f 1 and n i 2, 3, 4, . . . . The energylevel diagram for hydrogen shown in Active Figure 28.7 indicates the origin of thespectral lines described previously. The transitions between levels are representedby vertical arrows. Note that whenever a transition occurs between a state designatedby n i to one designated by n f (where n i n f ), a photon with a frequency(E i E f )/h is emitted. This can be interpreted as follows: the lines in the visiblepart of the hydrogen spectrum arise when the electron jumps from the third,fourth, or even higher orbit to the second orbit. Likewise, the lines of the Lymanseries (in the ultraviolet) arise when the electron jumps from the second, third, oreven higher orbit to the innermost (n f 1) orbit. Hence, the Bohr theory successfullypredicts the wavelengths of all the observed spectral lines of hydrogen.TIP 28.1 Energy DependsOn n Only for HydrogenAccording to Equation 28.13, the energydepends only on the quantumnumber n. Note that this is only truefor the hydrogen atom. For morecomplicated atoms, the energy levelsdepend primarily on n, but also onother quantum numbers.INTERACTIVE EXAMPLE 28.1The Balmer Series for HydrogenGoal Calculate the wavelength, frequency, and energy of a photon emitted during anelectron transition in an atom.Problem The Balmer series for the hydrogen atom corresponds to electronic transitionsthat terminate in the state with quantum number n 2, as shown in Figure 28.8.(a) Find the longest-wavelength photon emitted in the Balmer series and determine itsfrequency and energy. (b) Find the shortest-wavelength photon emitted in the sameseries.Strategy This is a matter of substituting values into Equation 28.17. The frequency canthen be obtained from c f and the energy from E hf. The longest wavelengthphoton corresponds to the one that is emitted when the electron jumps from the n i 3state to the n f 2 state. The shortest wavelength photon corresponds to the one that isemitted when the electron jumps from n i to the state n f 2.n∞6 –0.385–0.544–0.853–1.512BalmerseriesE (eV)0.00–3.40Figure 28.8 (Example 28.1)Transitions responsible for theBalmer series for the hydrogenatom. All transitions terminateat the n 2 level.Solution(a) Find the longest wavelength photon emitted in theBalmer series, and determine its energy.Substitute into Equation 28.17, with n i 3 and n f 2:Take the reciprocal and substitute, finding the wavelength:1 R H 1n 2 f 365R H 656.3 nm 1ni 2 R H 12 2 1 2 3 5R H36365(1.097 10 7 m 1 ) 6.563 107 mNow use c f to obtain the frequency:f c 2.998 108 m/s6.563 10 7 m4.568 10 14 HzCalculate the photon’s energy by substituting into Equation27.5:(b) Find the shortest wavelength photon emitted in theBalmer series.Substitute into Equation 28.17, with n i and n f 2.Take the reciprocal and substitute, finding the wavelength:E hf (6.626 10 34 J s)(4.568 10 14 Hz) 3.027 10 19 J 1.892 eV1 R H 1n 2 f 4R H 364.6 nm 1n i 2 R H 12 2 1 R H44(1.097 10 7 m 1 ) 3.646 107 m