A) \[6561\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[5464\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[800\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[4840\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: A
Solution :
Using relation \[\frac{1}{\lambda }=R\left( \frac{1}{{{n}_{1}}^{2}{{n}_{2}}^{2}} \right)\] For Lyman series \[{{n}_{1}}=1,\,{{n}_{2}}=2\] So, \[\frac{1}{{{\lambda }_{L}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)=\frac{3R}{4}\] ??(1) For \[{{H}_{\alpha }}\] line \[{{n}_{1}}=2,{{n}_{2}}=3\] \[\frac{1}{{{\lambda }_{H\alpha }}}=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)=\frac{5R}{36}\] ?..(2) From Eqs. (1) and (2), \[\frac{{{\lambda }_{H\alpha }}}{{{\lambda }_{L}}}=\frac{3}{4}\times \frac{36}{5}\] \[=\frac{27}{5}\] \[{{\lambda }_{H\alpha }}=\frac{27}{5}\times 1215\] \[=6561\overset{\text{o}}{\mathop{\text{A}}}\,\]
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