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