{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1191_Physics ProblemsTechnical Physics

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
532 Atomic Physics P42.43 Following Example 42.9 E γ = = × = × 3 4 42 1 13 6 1 71 10 2 74 10 2 4 15 a f a f . . . eV eV J f = × 4 14 10 18 . Hz and λ = 0 072 5 . nm . P42.44 E hc = = = λ λ λ 1 240 1 240 eV nm keV nm . For λ 1 0 018 5 = . nm , E = 67 11 . keV λ 2 0 020 9 = . nm , E = 59 4 . keV λ 3 0 021 5 = . nm , E = 57 7 . keV 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 = 2 39 . . 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 Z Z Z Z Z Z Z λ = − + × × × × = + + + F H G I K J × = F H G I K J 13 6 9 3 13 6 1 1 6 626 10 3 00 10 0 152 10 1 60 10 13 6 9 18 9 81 9 2 1 8 17 10 13 6 8 9 8 2 2 2 2 34 8 9 19 2 2 3 2 . . . . . . . . . a f a
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}