InterviewSolution
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InterviewSolution
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For a cartain metal, the `K` obsorption edge is at `0.72 Å`. The wavelength of `K_(alpha), K_(beta). And K_(gamma)` lines of of `K` series are `0.210 Å , 0.192 Å, and 0.180 Å, respectively. The eneggies of `K, L and M` orbit are `E_(K), E_(L) and E_(M)`, respectively. ThenA. `E_(K) = - 13.04 keV`B. `E_(L) = - 7.52 keV`C. `E_(M) = - 3.21 keV`D. `E_(K) = - 13.04 keV` |
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Answer» Correct Answer - A::B::C Energy of `K` absorption edge `E_(K) = (1242 eV nm)/(0.0172 nm) = 72.21 xx 10^(3) eV = 72.21 KeV` Energy of `K_(alpha)` line is `E_(K_alpha) = (hc)/(e lambda_(alpha)) = (1242 eV nm)/(0.021 nm) = 59.14 KeV` Similarly, `E_(K_beta) = (1242)/(0.0192) = 64.69 KeV` `E_(K_gamma) = (1242)/(0.0180) = 69 KeV` Energy of `K` shell ` = (E_(K_alpha) - E_(K))` `= (59.14 - 72.21) KeV = - 13.04 KeV` Energy of `L` shell ` = E_(K_beta) - 72.21 KeV` `= 64.69 KeV - 72.21 KeV = - 7.52 KeV` Energy of `M` shell ` = (E_(K_gamma) - E_(K)) = (1242 eV nm)/(0.018 nm) - 72.21 KeV` `= 69 KeV - 72.21KeV = - 3.21 KeV` |
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