What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?
Shane Macguire , 10 Years ago
Grade upto college level
1 Answers
Deepak Patra
Last Activity: 10 Years ago
λ = 0.10 nm = 10^10 m ; h = 6.63 * 10^34 J-s
C = 3 * 10^8 m/s ; e = 1.6 * 10^19 C
λ base min = hc/eV or V = hc/eλ
= 6.63 *10^-34 * 3 * 10^8/1.6 *10^-19 *10^-10 = 12.43 *10^3 V = 12.4 KV.
Max. Energy = hc/λ = 6.63* 10^-34 *3* 10^8/10^10 = 19.89 * 10^-18 = 1.989 * 10^-15 =2 * 10^-15 J.
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