In a Coolidge tube, the potential difference used to accelerate the electron is increased from 24 . 8 kV to 49 . 6 kV . As a result, the difference between the wavelength of K α -line and minimum wavelength becomes two times. Then the wavelength of K α -line is h c e = 12 . 4 k V A ∘

Step-by-step solution

Let ∆ E be the energy of k α-line then h c λ k α = Δ E k α ⇒ λ k α = h c Δ E k α Now λ 1  m i n = h c e × 24.8  k V = 12.4 24.8   A ∘ = 1 2   A ∘ and λ 2  m i n = h c e × 49.6  k V = 12.4 49.6   A ∘ = 1 4   A ∘ ⇒ 2 λ k α - 1 2 = λ k α - 1 4 ⇒ 2 λ k α - 1 = λ k α - 1 4 ⇒ λ k α = 1 - 1 4 ⇒ λ k α = 0 . 75 &