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

Options

A. 28
B. 29
C. 65
D. 66

Answer

B

Explanation

The cut-off wavelength when `V = V_(1) = 10 kV` is
`lambda_(1) = (h c)/(e V_(1)) = 1243.125 xx 10^(-13) m`
The cut-off wavelength when `V = V_(2) = 20 kV` is
`lambda_(2) = (h c)/(e V_(2)) = 621.56 xx 10^(-13) m`
The wavelength corresponding to `K_(alpha)` line is ,
`(1)/(lambda) = (3R)/(4) (Z - 1)^(2)`
From given information, `(lambda - lambda_(2)) = 3 (lambda - lambda_(1))`
Solving above equation, we get `Z = 29`