A) \[1:3\]
B) \[1:30\]
C) \[7:50\]
D) \[7:108\]
Correct Answer: D
Solution :
For first line of Lyman series, \[{{n}_{1}}=1\] and\[{{n}_{2}}=2\] \[\therefore \] \[\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] \[=R\left( 1-\frac{1}{4} \right)\] \[=\frac{3R}{4}\] For first line of Paschen series \[{{n}_{1}}=3\]and\[{{n}_{2}}=4\] \[\therefore \] \[\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[=R\left( \frac{1}{9}-\frac{1}{16} \right)\] \[=\frac{7R}{144}\] \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{7R}{144}\times \frac{4}{3R}\] \[=\frac{7}{108}\]
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