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When the voltage applied to an X-ray tube increased from V_(1)=10kV to V_(2)=20kV, the wavelength interval between the k_(alpha) line and the short-wave cut-off of the continuous X-rays spectrum increases by a factor n=3.0. Find the atomic number of the element of which the tube's anticathode is made. |
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Answer» SOLUTION :SUPPOSE `lambda_(0)=` wavelength of the characteristic `X`-rays line. Then using the FORMULA for short wavelength limit of limit of continous radiation `(lambda_(0)-(2piħc)/(eV_(1)))/(lambda_(0)-(2piħc)/(eV_(2)))=(1)/(N)` Hence `lambda_(0)=(2 piħc)/(eV_(1))((n-(V_(1))/(V_(2))))/(n-1)` Using also Moseley's law, we get `Ƶ=1+sqrt((8pic)/(3R lambda))=1+2sqrt((n-1)/(3 ħR_(n)-(V_(1))/(V_(2))))=29`. |
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