# The Shortest Wavelength Of He + Ion In Balmer Series Is X, Then Longest Wavelength In The Paschene Series Of Li +2 Is:

The shortest wavelength of Helium ion in Balmer Series is X

$$\frac{1}{\lambda He} = R_{H}(\frac{1}{2^{2}} – \frac{1}{\infty ^{2}})2^{2}$$ $$\Rightarrow \frac{1}{\lambda He} = \frac{R_{H}}{4} * 4$$ $$\Rightarrow \frac{1}{\lambda He} = \frac{R_{H}}{4}$$ $$\Rightarrow \frac{1}{R_{H}} = \lambda He$$ $$\Rightarrow x$$

The longest wavelength of Lithium-ion in the Paschene series is

$$\frac{1}{\lambda H} = R_{H}3^{2}(\frac{1}{3^{2}} – \frac{1}{4^{2}})$$ $$\Rightarrow \frac{1}{\lambda Li} = R_{H}(\frac{7}{144})$$ $$\Rightarrow \lambda _{4} = \frac{16}{7R_{H}}$$ $$\Rightarrow \frac{16}{7}x$$

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