A) \[911\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ }1215\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[1011\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ }1515\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[711\overset{\text{o}}{\mathop{\text{A}}}\,,\text{ }575\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) None of these
Correct Answer: A
Solution :
For Lyman series \[\frac{1}{\pi }=R\left[ \frac{1}{{{\left( 1 \right)}^{2}}}-\frac{1}{{{n}^{2}}} \right]\]Where, n = 2,3,4,5,? \[\frac{1}{{{\lambda }_{shortest}}}=R\left[ \frac{1}{{{\left( 1 \right)}^{2}}}-\frac{1}{\infty } \right]=10967700/s\] \[\frac{1}{{{\lambda }_{shortest}}}=R\frac{1}{10967700}=911\overset{0}{\mathop{\text{A}}}\,\] Longest wavelength limit \[\frac{1}{{{\lambda }_{shortest}}}=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]=\frac{3}{4}R\] \[{{\lambda }_{longets}}=\frac{4}{3R}\] \[=\frac{4}{3\times 10967700}=1215\overset{0}{\mathop{\text{A}}}\,\]
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