A) \[n=1\]
B) \[n=2\]
C) \[n=3\]
D) \[n=\infty \]
Correct Answer: B
Solution :
When an atom comes down from some higher energy level to the second energy level\[({{n}_{1}}=2)\]and \[{{n}_{2}}=3,4,5,\text{ }...\]), then the lines of the spectrum are obtained in the visible part. \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right);\]where \[n=3,4,5,.....\] The shortest wavelength of the series corresponds to\[n=\infty \]is\[3646\overset{o}{\mathop{\text{A}}}\,\].
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