The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Explanation

`lambda _(min) = (ch)/(e V_(1))`
and Hence `lambda _(1) = (ch)/(e V_(1))` and `lambda _(2) = (ch)/(e V_(2))`
or `(ch)/(e) [(1)/(V_(2)) - (1)/(V_(1))] = lambda _(2) - lambda _(1) = Delta lambda = 26 xx 10^(-12)`
or `(ch)/(e) [(1)/(3 V_(1)) - (1)/(V_(1))] = 26 xx 10^(-12) (:.V_(2) = 1.5 e V_(1))`
Solving for `V_(1)`, we get `V_(1) = 16000 V or 16 kV`