Quantum Physics

908 Chapter 28 Atomic PhysicsThe orbit with the smallest radius, called the Bohr radius, a 0 , corresponds ton 1 and has the value4a 0– ea 0+ e9a 0ACTIVE FIGURE 28.6The first three circular orbitspredicted by the Bohr model ofthe hydrogen atom.Log into PhysicsNow atwww.cp7e.com and go to ActiveFigure 28.6, where you can choosethe initial and final states of thehydrogen atom and observe thetransition.ENERGYn∞54321BalmerseriesLymanseriesE (eV)0.00–0.5442–0.8504–1.512Paschenseries–3.401–13.606ACTIVE FIGURE 28.7An energy level diagram for hydrogen.Quantum numbers are given on theleft and energies (in electron volts)are given on the right. Vertical arrowsrepresent the four lowest-energytransitions for each of the spectralseries shown. The colored arrows forthe Balmer series indicate that thisseries results in visible light.Log into PhysicsNow atwww.cp7e.com and go to ActiveFigure 28.7, where you can choosethe initial and final states of thehydrogen atom and observe thetransition.a 0 [28.10]mk e e2 0.052 9 nmA general expression for the radius of any orbit in the hydrogen atom is obtainedby substituting Equation 28.10 into Equation 28.9:r n n 2 a 0 n 2 (0.052 9 nm) [28.11]The first three Bohr orbits for hydrogen are shown in Active Figure 28.6.Equation 28.9 may be substituted into Equation 28.8 to give the following expressionfor the energies of the quantum states:E n m ek 2 e e 4 1 2 n 1, 2, 3, . . . [28.12]2 2 nIf we insert numerical values into Equation 28.12, we obtainE n 13.6[28.13]n 2 eVThe lowest energy state, or ground state, corresponds to n 1 and has an energyE 1 m e k 2 e e 4 /2 2 13.6 eV. The next state, corresponding to n 2, has anenergy E 2 E 1 /4 3.40 eV, and so on. An energy level diagram showing theenergies of these stationary states and the corresponding quantum numbers isgiven in Active Figure 28.7. The uppermost level shown, corresponding to E 0and n : , represents the state for which the electron is completely removedfrom the atom. In this state, the electron’s KE and PE are both zero, which meansthat the electron is at rest infinitely far away from the proton. The minimum energyrequired to ionize the atom—that is, to completely remove the electron—iscalled the ionization energy. The ionization energy for hydrogen is 13.6 eV.Equations 28.3 and 28.12 and the third Bohr postulate show that if the electronjumps from one orbit with quantum number n i to a second orbit with quantumnumber, n f , it emits a photon of frequency f given byf E i E f m ek 2 e e 42 [28.14]hn 1f n i2where n f n i .Finally, to compare this result with the empirical formulas for the various spectralseries, we use Equation 28.14 and the fact that for light, f c, to get1 f c m ek e 2 e 44c 3 1n f2 1n i2[28.15]A comparison of this result with Equation 28.1 gives the following expression forthe Rydberg constant:R H m ek 2 e e 4[28.16]4c 3If we insert the known values of m e , k e , e, c, and into this expression, the resultingtheoretical value for R H is found to be in excellent agreement with the value determinedexperimentally for the Rydberg constant. When Bohr demonstrated thisagreement, it was recognized as a major accomplishment of his theory.In order to compare Equation 28.15 with spectroscopic data, it is convenient toexpress it in the form124 3 1 R H 1n f2 1n i2[28.17]