If the wavelength of 1^ st line of Balmer series of hydrogen is 6561 Å , the wavelength of the 2^ nd line of series will be

Step-by-step solution

For the first line of Balmer series, 1 ₁ =R ( 1 2^2 - 1 3^2 )= 5 R 36 For the second line of Balmer series. 1 ₂ =R ( 1 2^2 - 1 4^2 )= 3 R 16 ₂ ₁ = 5 R / 36 3 R / 16 = 20 27 ; or ₂= 20 27 (6561 Å)=4860 Å