The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number 'Z'. The value of 'Z' is

Step-by-step solution

The wavelength of a spectral line is given by Rydberg's formula: 1 = R Z^2 ( 1 n₁^2 - 1 n₂^2 ) For the shortest wavelength in the Balmer series of hydrogen ( Z = 1 ), n₁ = 2 and n₂ = : 1 ₁ = R (1)^2 ( 1 2^2 - 0 ) = R 4 For the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number Z , n₁ = 4 and n₂ = : 1 ₂ = R Z^2 ( 1 4^2 - 0 ) = R Z^2 16 Given ₁ = ₂ , we have: R 4 = R Z^2 16 Z^2 = 4 Z = 2 Answer: 2