A) \[94.21nm\]
B) \[91.16nm\]
C) \[911.6nm\]
D) \[933.6nm\]
Correct Answer: B
Solution :
For Lyman series,\[\frac{1}{\lambda }=R\left( \frac{1}{{{(1)}^{2}}}-\frac{1}{{{n}^{2}}} \right)\]hence, for \[\lambda \] to be smallest, \[n\] should be greatest (i.e.,\[\infty \]). \[\therefore \] \[\frac{1}{{{\lambda }_{\min }}}=1.097\times {{10}^{-2}}\left( 1-\frac{1}{{{\infty }^{2}}} \right)\] \[=1.097\times {{10}^{-2}}\] or \[{{\lambda }_{\min }}=\frac{1}{1.097\times {{10}^{-2}}}\] \[=91.16\,\,nm\]
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