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Find the voltage applied to an X-rays tube with nickel anticathode if the wavelength difference between the k_(alpha) line and the short-wave cut-off of the continuous X-rays spectrum is equal to 84 p m. |
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Answer» Solution :From Mosley's LAW `lambda k_(alpha)(Ni)=(8 pi c)/(3R) (1)/((Ƶ-1)^(2))` where `Ƶ= 28` for `Ni`. SUBSTITUTION gives `lambda k_(alpha) (Ni)=166.5 p m` Now the short WAVE cut of off the continuous spectrum MUST be more energetic (smaller wavelength) otherwise `k_(alpha)` lines will not emerge. Then since `Delta lambda= lambda k_(alpha)-lambda_(0)=84 p m` we get `lambda_(0)=82.5 p m` This corresponds to voltage of `V=(2piħ c)/(e lambda_(0))` Substitution gives `V=15.0 kV`. |
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