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Chemistry Question on Structure of atom

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v=3.29×1015Hz(132−1n2)v=3.29\times10^{15} Hz \left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) Calculate the value of n if the transition is observed as 1285 nm.

A

3

B

4

C

5

D

6

Answer

5

Explanation

Solution

v=3.29×1015(132−1n2)v=3.29\times10^{15} \left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) v=cλ=3×1081285×10−9=3.29×1015(132−1n2)v=\frac{c}{\lambda}=\frac{3\times10^{8}}{1285\times10^{-9}}=3.29\times10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) 1n2=19−3×1081285×10−9×13.29×1015⇒n2=25⇒n=5\frac{1}{n^{2}}=\frac{1}{9}-\frac{3\times10^{8}}{1285\times10^{-9}}\times\frac{1}{3.29\times10^{15}} \Rightarrow n^{2}=25 \Rightarrow n=5 The radiation corresponding to 1285 nm lies in the infrared region.