If the first line of Lyman series has a wavelength 1215.4 Å , the first line of Balmer series is approximately

Step-by-step solution

From hydrogen spectrum, when electron transist from n₂ orbit to n₁ orbit, the emitted wavelength is given by the formula 1 _L =R Z^2 ( 1 n₁^2 - 1 n₂^2 ) Here, R= Rydberg constant. For Lyman series, n₁=1, n₂=2 For First Lyman line, 1 _L =R Z^2 ( 1 1^2 - 1 2^2 ) 1 _L = 3 4 R Z^2 For Balmer series, n₁=2, n₂=3,4 For first balmer line, 1 _B =R Z^2 ( 1 2^2 - 1 3^2 )=R Z^2 5 36 Dividing Eqs. (i) by (ii), we get 1 / _L 1 / _B = 3 / 4 R Z^2 5 / 36 R Z^2 = 3 4 36 5 = 27 5 array llll & _B _L = 27 5 Here, & _L & =1215.4 Å & _B