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QJEEM19Chf0e7: JEE Main 2019 Chemistry - Class 11 JEEM19Chf0e7 Solution: The ratio of the shortest wavelength of two spectral series of…

The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are :

Multiple Choice Options

  • A Paschen and Pfund
  • B Balmer and Brackett
  • C Lyman and Paschen
  • D Brackett and Pfund

Step-by-Step Verified Solution

Shortest wavelength can found when n2 = ∞\infty

λshortest=RZ2{1n12−1∞2}{\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}

Here n1 = series number.

⇒\Rightarrow λshortest={\lambda _{shortest}} = RZ2n12{{{R{Z^2}} \over {n_1^2}}} = R(1)2n12{{{R{{\left( 1 \right)}^2}} \over {n_1^2}}}

∴\therefore λAλB=(nBnA)2{{{\lambda _A}} \over {{\lambda _B}}} = {\left( {{{{n_B}} \over {{n_A}}}} \right)^2}

Given, (nBnA)2=9{\left( {{{{n_B}} \over {{n_A}}}} \right)^2} = 9

⇒\Rightarrow nBnA=3{{{n_B}} \over {{n_A}}} = 3

∴\therefore Series number of B = 3 ×\times Series number of A

So,

If series number of A is Lyman( n = 1 )

then series number of B is Paschen( n = 3 ).

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