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What metal has in its absorption spectrum the difference between the frequencies of X-rays K and L absorption edges equal to Delta omega= 6.85.10^(18) S^(-1) ? |
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Answer» Solution :The difference in FREQUENCIES of the `K` and `L` absorption edges is equal, accorfing to the Bohr picture, to the frequency of the `k_(alpha)` line(SEE the diagram below). Thus by Mosely's formule `DELTA omega =(3)/(4)R(Ƶ-1)^(2)` or `Ƶ=1+sqrt((4Delta omega)/(3R))=22`  The metal is titanium. |
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