1191_Physics ProblemsTechnical Physics

1191_Physics ProblemsTechnical Physics - 532 P42.43 Atomic...

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532 Atomic Physics P42.43 Following Example 42.9 E γ =− = × = × 3 4 42 1 13 6 1 71 10 2 74 10 2 41 5 af a f .. . eV eV J f 414 10 18 . H z and λ = 00725 n m . P42.44 E hc == = λλ 1240 eV nm keV nm . For 1 00185 = n m , E = 67 11 k e V 2 00209 = n m , E = 59 4 . k eV 3 00215 = n m , E = 57 7 The ionization energy for the K shell is 69.5 keV, so the ionization energies for the other shells are: FIG. P42.44 L shell keV = 11 8 . M shell keV = 10 1 . N shell keV = 239 . . P42.45 The K β x-rays are emitted when there is a vacancy in the ( n = 1) K shell and an electron from the ( n = 3 ) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and by one in its final state. hc ZZ Z + ×⋅× ×× + + + F H G I K J ×= F H G I K J −− 13 6 9 3 13 6 1 1 6626 10 300 10 0152 10 160 10 13 6 9 18 9 81 9 21 817 10 136 8 9 8 2 2 2 2 34 8 91 9 2 2 3 2 . ej e j e j a f eV eV Js ms m J e V eV
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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