A) \[\frac{5A}{9}\]
B) \[\frac{36A}{7}\]
C) \[\frac{36A}{5}\]
D) \[\frac{9A}{5}\]
Correct Answer: B
Solution :
Shortest wavelength is corresponding to best ine \[\therefore \]\[{{n}_{L}}=1\] (Lyman series) \[{{n}_{H}}=\infty \](infinite) \[\frac{1}{A}=r\times {{(1)}^{2}}\left\{ \frac{1}{12}-\frac{1}{2} \right\}=R\] Longest wavelength \[\equiv \] 1st Line \[\therefore \] = 3 \[{{n}_{H}}=4\] \[\frac{1}{\lambda }=r\times {{(2)}^{2}}\left\{ \frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}} \right\}=\frac{r\times 7}{36}\]\[\lambda =\frac{36A}{7}\]
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