A) \[\frac{9x}{5}\]
B) \[\frac{36x}{5}\]
C) \[\frac{x}{4}\]
D) \[\frac{5x}{9}\]
Correct Answer: A
Solution :
\[\frac{1}{\lambda }\,\,=\,\,{{R}_{H}}\,\times {{Z}^{2}}\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{2_{2}^{2}} \right]\] \[\frac{1}{x}\,\,=\,\,{{R}_{H}}\,\times 1\,\,\left[ \frac{1}{1}-\frac{1}{\infty } \right]\] \[x\,\,=\,\,\frac{1}{{{R}_{H}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\,\,\,\,{{R}_{H}}\,\,=\,\,\frac{1}{x}\] \[\frac{1}{\lambda }\,\,=\,\,{{R}_{H}}\times 4\,\left[ \frac{1}{4}-\frac{1}{9} \right]\,\] \[\frac{1}{\lambda }\,\,=\,\,{{R}_{H}}\times 4\,\,\times \,\,\frac{5}{36}\,\] \[\,\frac{1}{\lambda }\,\,=\,\,\frac{5{{R}_{H}}}{9}\] \[\,x\,\,=\,\,\frac{9}{5}\,\,\times \,\,x\]
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