A) 91 nm
B) 192nm
C) 406 nm
D) \[9.1\times {{10}^{-8}}nm\]
Correct Answer: A
Solution :
\[\frac{1}{\lambda }={{\overline{v}}_{H}}={{\overline{R}}_{H}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[=1.097\times {{10}^{7}}\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]\] \[\therefore \]\[\lambda =\frac{1}{1.097\times {{10}^{7}}}m=9.11\times {{10}^{-8}}m\] \[=91.1\times {{10}^{-9}}m=91.1nm\] \[(1nm={{10}^{-9}}m)\]
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