Question

What is the shortest wavelength present in the Paschen series of spectral lines?

Answer 1

Paschen series is come out when we study about the hydrogen atom
paschen series is 1 of 5 in hydrogen atom
paschen series is 3rd series series of hydrogen atom spectra
it belong to IR region
the shortest wavelength in transition between 8>3
the wavelength of this line is near by 955nm
it prrsent into IR region which os not see by naked eyes.
hope it u
further knowledge u can msg me

Answer 2

$\Large{\green{\underline{\underline{\sf{\blue{Solution:}}}}}}$

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Rydberg's formula is given as,

$\longrightarrow$ $\tt \dfrac{hc}{\lambda}\:=\:21.76\times 10^{-19}\left[\dfrac{1}{n_1^2}\,-\, \dfrac{1}{n_2^2}\right]$

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Where,

$\hookrightarrow$ $\texttt{h = planks constant =}$$\tt 6.6\times 10^{-34}Js$

$\hookrightarrow$ $\texttt{c = Speed of light =}$ $\tt 3\times 10^8ms^{-1}$

( $\tt n_1$ and $\tt n_2$ are integers )

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$\longrightarrow$ The shortest wavelength present in the Paschen series of the spectral lines is given for values $\tt n_1\:=\:3\:$ and $\tt \:n_2\:=\: \infty$

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$\tt \dfrac{hc}{\lambda}\:=\:21.76\times 10^{-19}\left[\dfrac{1}{(3)^2}\,-\, \dfrac{1}{(\infty)^2}\right]$

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$\tt \lambda\:=\: \dfrac{6.6\times 10^{-34}\times 3\times 10^8\times 9}{21.76\times 10^{-19}}$

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$\tt \lambda\:=\:8.189\times 10^{-7}m$

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$\tt \lambda\:=\:8.189nm$

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$\hookrightarrow$ Therefore the shortest wavelength present in the Paschen series is $\tt{\orange{8.189nm}}$