JEE MAIN - Chemistry (2019 - 10th April Evening Slot - No. 12)
The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The
spectral series are :
Paschen and Pfund
Balmer and Brackett
Lyman and Paschen
Brackett and Pfund
Explanation
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 ).
$${\lambda _{shortest}} = R{Z^2}\left\{ {{1 \over {n_1^2}} - {1 \over {{\infty ^2}}}} \right\}$$
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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