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