A) Balmer series
B) Lyman series
C) Paschen series
D) Pfund series (Rydberg constant \[R=1.097\times {{10}^{7}}\]per metre)
Correct Answer: C
Solution :
\[\frac{1}{\lambda }=R\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\Rightarrow \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}=\frac{1}{R\lambda }\] \[=\frac{1}{1.097\times {{10}^{7}}\times 18752\times {{10}^{-10}}}\]\[=0.0486=\frac{7}{144}.\] But \[\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}=\frac{7}{144}\Rightarrow {{n}_{1}}=3\] and n2 = 4 (Paschen series)
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