1199_Physics ProblemsTechnical Physics

1199_Physics ProblemsTechnical Physics - 540 Atomic Physics...

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540 Atomic Physics Therefore we require ..... =− F H G I K J + F H G I K J = 4 6 0 00 2 r a r a with solutions ra 35 0 ej . We substitute the last two roots into Pr af to determine the most probable value: When raa = 0 7 6 3 9 ., a = 00519 0 . . When a =+ = 5 2 3 6 a a f = 0191 0 . . Therefore, the most probable value of r is 5 2 3 6 += aa .. (b) Prdr r a r a ed r 0 2 0 3 0 2 0 1 8 2 0 zz = F H G I K J F H G I K J Let u r a = 0 , dr a du = 0 , P r dr u u u e dr u u u e du u u u e uu u e j e j 0 22 0 432 0 42 0 1 8 44 1 8 1 8 48 8 1 z + = + = + + + = This is as required for normalization. P42.66 E hc E == = λλ 1240 eV nm λ 1 310 = nm, so E 1 400 = . e V 2 400 = E 2 310 = e V 3 1378 = E 3 0900 = e V and the ionization energy = 410 e V . FIG. P42.66 The energy level diagram having the fewest levels and consistent with these energies is shown at
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