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Wave Optics

Question
CBSEENPH12039941

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

  • 4/9

  • 9/4

  • 27/5

  • 5/27

Solution

D.

5/27

In hydrogen atom, wavelength of characteristic spectrum

1 over straight lambda space equals space Rz squared open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets

For space layman space series space straight n subscript 1 space equals space 1 comma space straight n subscript 2 space equals space 2
1 over straight lambda subscript 1 space equals space Rz squared open square brackets 1 over 1 squared minus 1 over 2 squared close square brackets space space... space left parenthesis straight i right parenthesis space

For space Balmer space series space straight n subscript 1 space equals space 2 comma space straight n subscript 2 space equals space 3

1 over straight lambda subscript 2 space equals space Rz squared open square brackets 1 over 2 squared minus 1 over 3 squared close square brackets space space... space left parenthesis ii right parenthesis space

Dividing space eq space left parenthesis ii right parenthesis space by space eq space left parenthesis straight i right parenthesis space space we space get
straight lambda subscript 1 over straight lambda subscript 2 space equals space fraction numerator Rz squared open square brackets 1 fourth minus 1 over 9 close square brackets space over denominator Rz squared open square brackets 1 over 1 minus 1 fourth close square brackets space end fraction equals fraction numerator begin display style 5 over 36 end style over denominator begin display style 3 over 4 end style end fraction
straight lambda subscript 1 over straight lambda subscript 2 space equals space 5 over 36 straight x space 4 over 3 space equals space 5 over 27