The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Step-by-step solution

For hydrogen or hydrogen type atoms 1 =R Z^2 ( 1 n_f^2 - 1 n_i^2 ) In the transition from n_i n_f aligned & 1 Z^2 ( 1 n_f^2 - 1 n_i^2 ) & ₂ ₁ = Z₁^2 ( 1 n_f^2 - 1 n_i^2 )₁ Z₂^2 ( 1 n_f^2 - 1 n_i^2 )₂ & aligned ₂= ₁ Z₁^2 ( 1 n_f^2 - 1 n_i^2 )₁ Z₂^2 ( 1 n_f^2 - 1 n_i^2 )₂ Substituting the values, we have = (6561 Å)(1)^2 ( 1 2^2 - 1 3^2 ) (2)^2 ( 1 2^2 - 1 4^2 ) =1215 Å Correct option is (a). Analysis of Question (i) Question is simple. (ii) In modern physics, mostly questions are asked on the emission of photon by th