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Proceeding from Mosely's law find: (a) the wavelength of the k_(alpha) line in aluminum and cobalt: (b)the difference in binding energies of K and L electrons in Vanadium. |
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Answer» Solution :(a) From Moselety's law `omega_(k alpha)=(3)/(4)R(Ƶ-sigma)^(2)` or `lambda k_(alpha)=(2 pi Ƶc)/(omegak_(alpha))=(8pi c)/(3R)(1)/((Ƶ-sigma)^(2))` We shall take `sigma=1`. For Aluminum `(Ƶ=13)` `lambdak_(alpha)(AL)=843.2 p m` and for cobalt `(Ƶ=27)` `lambdak_(alpha)(Co)=179.6p m` (b) This difference is nearly equal to the energy of the `k_(alpha)` LINE which by Moseley's law is equal to `(Ƶ=23` for VANADIUM) `DeltaE=ħ onegak_(alpha)=(3)/(4)xx13.62xx22xx22=4.94 KEV` |
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