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?

Options

A. `~~ 10 kV`
B. `~~ 16 kV`
C. `~~ 50 kV`
D. `~~ 75 kV`

Answer

B

Explanation

`lambda_(min) = (hc)/(eV) rArr lambda_(1) (hc)/( e V_(1))` and `lambda_(2) = (hc)/( eV_(2))`
`:. Delta lambda = lambda_(2) - lambda_(1) = (hc)/(e ) [ (1)/(V_(2)) - (1)/(V_(1)) ]`. Given `V_(2) = 1.5 V_(1)`
On solving we get `V_(1) = 16000 volt = 16 kV`.