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?

For balmer series: 1/λ = R(1/n₁² - 1/n₂²) For first line of hydrogen: n₁ = 2 and n₂ = 3
1/λH = R(1/4 - 1/9) = 5R/36
For second line of Helium: n₁ = 2 and n₂ = 4, Z = 2
1/λHe = Z²R(1/4 - 1/16) = 4R(3/16) = 3R/4
Now, λH/λHe = (5R/36)/(3R/4) = 5/27
λHe = 27λH/5 = 27 × 6561/5 = 3542.94 Å ≈ 1215 Å