Emission transitions in the Paschen series end at orbit $n$ $=$ $3$ and start from orbit $n$ and can be represented as $v = 3.29 \times 10^{15} (H z) [1 / 3^{2} - 1 / n^{2}]$ Calculate the value of $n$ if the transition is observed at 1.285 nm. Find the region of the spectrum.

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Solution

$υ = c / λ = \frac{3 \times 10^{8}}{1285 \times 10^{- 9}}$
and also $υ = 3.39 \times 10^{15} (1 / 3^{2} - 1 / n^{2})$
$1 / n^{2} = 1 / 9 - \frac{3 \times 10^{8}}{1285 \times 10^{- 9}} \times \frac{1}{3.29 \times 10^{15}}$
= 0.111 - 0.071
= 0.04 or 1/25
or $n^{2}$ = 25
or $n = 5$
and the spectrum lies in the infrared region.

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Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v=3.29×1015(Hz)[1/32−1/n2] Calculate the value of n if the transition is observed at 1.285 nm. Find the region of the spectrum.

υ=c/λ=3×1081285×10−9 and also υ=3.39×1015(1/32−1/n2)1/n2=1/9−3×1081285×10−9×13.29×1015 = 0.111 - 0.071 = 0.04 or 1/25 or n2 = 25 or n=5 and the spectrum lies in the infrared region.

Emission transitions in the Paschen series end at orbit $n$ $=$ $3$ and start from orbit $n$ and can be represented as $v = 3.29 \times 10^{15} (H z) [1 / 3^{2} - 1 / n^{2}]$ Calculate the value of $n$ if the transition is observed at 1.285 nm. Find the region of the spectrum.