In a Coolidge tube, the potential difference used to accelerate the electrons 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. The initial wavelength of the K α -line is [Take h c e = 12 .4 kV 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 the cutoff wavelengths in the two cases are λ 1 m i n = h c e × 24.8  k V = 12.4 24.8 A ° = 1 2 A ° 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 = 3 4 Å