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

Solution

For the shortest wavelength in Paschen series,
n₂ =

\infty

and n₁ = 3

Using Rydberg's formula,

\frac{1}{λ} = R [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}]

where, the value of R is 1.097

\times

10⁷m^-1.

Therefore,

\frac{1}{λ_{\min}} = R (\frac{1}{3^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{9}

\Rightarrow

λ_{\min} = \frac{9}{R} = \frac{9}{1.097 \times 10^{7}} m

i.e.,

λ_{\min} = \frac{9 \times 10^{- 7} \times 10^{10}}{1.097} Å

= 8204.2 Å

, is the value of the shortest wavelength.

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