When the voltage applied to an X-ray tube increases from `10` to `20 kV` the wavelength difference between the ka line and the short wave cut-off of continuous X-ray spectrum increases by a factor of `3.0`. Identify the target element. Take `4/3R = 1200 Å` where R is Rydberg’s contant and `hc = 12.4 keV Å`

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

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 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 increased from V_(1)=15.5kV to V_(2)=31kV the wavelength interval between the K_(alpha) line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of 1.3 . If te atomic number of the element of the target is z. Then the value of (z)/(13) will be: (take hc=1240eVnm and R=1xx(10^(7))/(m))