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Question
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?
(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
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Solution
Given :-
Wavelength of the X-ray, `λ = 0.10 "nm"`
Planck's constant , `h = 6.63 xx 10^-34 "J-s"`
Speed of light, `c = 3 xx 10^8 "m/s"`
Minimum wavelength is given by
`lambda_min = (hc)/(eV)`
`⇒ V = (hc)/(elambda_min)`
`⇒ V = (6.63 xx 10^-34 xx 3 xx 10^8)/(1.6 xx 10^-19 xx 10^-10)`
`⇒V = 12.43 xx 10^3 "V" = 12.4 "kV"`
Maximum energy of the photon (E) is given by
`E = (hc)/lambda`
`⇒ E = (6.63 xx 10^-34 xx 3 xx 10^8)/(10^-10)`
`⇒ E = 19.89 xx 10^-16`
`⇒ E = 1.989 xx 10^-15 ≈ 2 xx 10^-15 "J"`
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