#### 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 × 10^{8} m/s.)

#### 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"`