A) \[1002.7\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[1215.67\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[1127.30\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) \[911.7\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
E) \[1234.7\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: D
Solution :
For Lyman series, \[{{n}_{1}}=1\] Hence, \[\overline{v}={{R}_{H}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)\] where, \[\overline{v}=\frac{1}{\lambda }\] For\[\lambda \]to be minimum (shortest),\[\overline{v}\]should be maximum. This can be so when\[{{n}_{2}}\]is maximum, ie, \[{{n}_{2}}=\infty \].Hence, \[\frac{1}{\lambda }=109678\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)\] Or \[\frac{1}{\lambda }=109678\] Or \[\lambda =911.7\times {{10}^{-10}}m\] \[=911.7\text{ }{\AA}\]
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