The wavelength of the second line of Balmer series is 486.4 ~nm . What is the wavelength of the first line of Lyman series?

Step-by-step solution

The Rydberg formula for the wavelength of a spectral line in a hydrogen-like atom is given by 1 = R ( 1 n₁^2 - 1 n₂^2 ) Z^2 . For hydrogen, Z = 1 . The second line of the Balmer series corresponds to the transition from n₂ = 4 to n₁ = 2 . Thus, 1 _B = R ( 1 2^2 - 1 4^2 ) = R ( 1 4 - 1 16 ) = R ( 3 16 ) . Given _B = 486.4 nm , we have R = 16 3 486.4 nm ⁻¹ . The first line of the Lyman series corresponds to the transition from n₂ = 2 to n₁ = 1 . Thus, 1 _L = R ( 1 1^2 - 1 2^2 ) = R ( 1 - 1 4 ) = R ( 3 4 ) . Substitut