In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

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Correct answer: D. (d) In hydrogen atom, wavelength of characteristic spectrum 1 =Rz^2( 1n_1^2- 1n_2^2) For Lyman series n_1=1 ,n_2=2 1 _1=Rz^2( 1 (1 )^2- 1 (2 )^2) ……….(i) For Balmer series n_1=2 ,n_2=3 1 _2=Rz^2( 1 (2 )^2- 1 (3 )^2) …………(ii) Dividing Eq. (ii) by Eq. (i) we get _1 _2= Rz^2( 14- 19)Rz^2(1- 14)= 536 34 _1 _2= 536 43= 527

$In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is$

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