when the voltage applied to an X-ray tube increases from `V_(1) = 10 kV` to `V_(2) = 20 kV`, the wavelenght interval between `K_(alpha)` line and cut-off wavelenght of continuous spectrum increases by a factor of `3`. Atomic number of the metallic target is

when the voltage applied to an X-ray tube increases from `V_(1) = 10 k

1.	when the voltage applied to an X-ray tube increases from `V_(1) = 10 kV` to `V_(2) = 20 kV`, the wavelenght interval between `K_(alpha)` line and cut-off wavelenght of continuous spectrum increases by a factor of `3`. Atomic number of the metallic target is
Answer» Correct Answer - 30 `lambda_(C_2) = (h)/(eV) = (12375)/(10 xx 10^(3)) Å = 1.2375 Å` (i) `lambda_(C_1) = 0.61875 Å` (ii) `(1)/(lambda_(k_alpha)) = (Z - 1)^(2) {(1)/(1) - (1)/(4)} xx 10^(7)` (iii) It is give `3 xx (lambda_(k_alpha) - 1.2375) = lambda_(k_alpha) - 0.61875` `implies lambda_(k_alpha) = 1.54338 Å` (iv) Putting this value in (iii). `Z - 1 = 29` `implies Z = 30`

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when the voltage applied to an X-ray tube increases from `V_(1) = 10 kV` to `V_(2) = 20 kV`, the wavelenght interval between `K_(alpha)` line and cut-off wavelenght of continuous spectrum increases by a factor of `3`. Atomic number of the metallic target is

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