A) \[\frac{R}{4}\]
B) \[\frac{3R}{4}\]
C) \[\frac{R}{2}\]
D) \[2R\]
Correct Answer: B
Solution :
Key Idea: When an electron in an atom jumps from higher orbits to the ground orbit, Lyman series is found. The wavelength of lines obtained in Lyman series is given by \[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}^{2}}} \right)\] \[n=2,3,4,.....\] For first line of Lyman series, \[n=2\] \[\therefore \] \[\bar{v}=\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right)\] or \[=R\left( \frac{4-1}{4} \right)\] or \[\bar{v}=\frac{3R}{4}\] Note: For shortest wavelength of Lyman series, put \[n=\infty .\]
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