A) \[1215\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[6560\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[8225\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[12850\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: C
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, \[\lambda =\frac{18800\times 7\times 9}{144}=8225\overset{\text{o}}{\mathop{\text{A}}}\,\]
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