A voltage applied to an `X`-ray tube being increased `eta = 1.5` times, the short-wave limit of an `X`-ray continuous spectrum shifts by `Delta lambda = 26 p m`. Find the initial voltage applied to the tube.
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The maximum energy given to X-ray photon is given by `hv_(max)= eV`, where V = applied voltage to X-ray tube `therefore` `V_(max)=(eV)/(h)` Therefore `lambda_(min)=(ch)/(eV)` and `lambda_(2)=(ch)/(eV_(2))` Therefore, `Deltalambda = lambda_(2)-lambda_(1)=(ch)/(e)[(1)/(V_(2))-(1)/(V_(2))]` Given `V_(2) =1.5 V_(1)` Solving we get `V_(1)= 516000 Volt`.
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