The wavelength of K base α X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of K base α X-ray?
Simran Bhatia , 11 Years ago
Grade 11
1 Answers
Aditi Chauhan
Last Activity: 11 Years ago
K base λ = 21.3 * 10^-12 pm, Now, E base k – E base l = 1242/21.3* 10^-3 = 58.309 kev
E base L = 11.3 kev, E base k = 58.309 + 11.3 = 69.609 kev
Now, Ve = 69.609 KeV, or V = 69.609 KV.
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