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  4. The first member of the paschen series in hydrogen spectrum of wave length is 18800 overset texto mathop textA The short Wavelength limit of Paschen series is:

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Q. The first member of the paschen series in hydrogen spectrum of wave length is $ 18800\overset{\text{o}}{\mathop{\text{A}}}\, $ The short Wavelength limit of Paschen series is:

EAMCETEAMCET 2000

Solution:

The wavelength of radiation emitted in, Paschen series $ \frac{1}{\lambda }=R\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{n}^{2}}} \right] $ For first member, n = 4 $ \therefore $ $ \frac{1}{18800}=R\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right] $ $ \therefore $ $ \frac{1}{18800}=R\times \left( \frac{16-9}{144} \right) $ or $ R=\frac{144}{18800\times 7} $ ?(i) For shortest wavelength limit (series limit) $ n=\infty $ $ \frac{1}{\lambda }=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{\infty }^{2}}} \right) $ $ =\frac{144}{18800\times 7}\times \frac{1}{9} $ [from Eq. (i)] Hence, $ \lambda =\frac{18800\times 7\times 9}{144}=8225\overset{\text{o}}{\mathop{\text{A}}}\, $