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

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

Accepted Answer

Shortest wavelength can found when n₂ =

\infty

{\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}

Here n₁ = series number.

\Rightarrow

{\lambda _{shortest}} =

{{{R{Z^2}} \over {n_1^2}}}

=

{{{R{{\left( 1 \right)}^2}} \over {n_1^2}}}

\therefore

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

Given,

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

\Rightarrow

{{{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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