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2.55    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}\ (\textup{Hz})\left[\frac{1}{3^2} - \frac{1}{n^2} \right ]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

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Given transition in the Paschen series end at orbit n=3 and starts from orbit n: v = 3.29\times 10^{15}\ (\textup{Hz})\left[\frac{1}{3^2} - \frac{1}{n^2} \right ]                          .........................(1)

\nu = \frac{c}{\lambda} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m}                                          .........................(2)

Equating both (1) and (2) equations: we get

3.29\times10^{15}\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m}

\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = \frac{3.0\times10^8m/s}{1285\times10^{-9}m\times3.29\times10^{15}}

\left [ \frac{1}{3^2}-\frac{1}{n^2} \right ] s^{-1} = 0.07096 s^{-1}

\frac{1}{9} - 0.07096 = \frac{1}{n^2}

\frac{1}{n^2} = 0.0401 \Rightarrow n^2 =25 \Rightarrow n=5

Therefore, the radiation corresponding to 1285 nm lies in the infrared region.

Posted by

Divya Prakash Singh

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