Question

The shortes wavelength of he atom in balmer series is x then longest wavelenth in the paschen series of li+2 is??;

Answer 1

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the shortest wavelength of He+ ion
lambda = α

1/α = 1/x
= RZ^2 (1/2^2 - 1/ §^2)
= R2^2(1/4)
= 1/x
Then, longest wavelength in Paschen series of line Li2+ will be from n = 4 to n = 3

1/α = RZ^2 (1/3^2 - 1/4^2)
= R7/6
R = 1/x
α = 7x/6