A) \[8.22\times {{10}^{14}}{{s}^{-1}}\]
B) \[3.29\times {{10}^{15}}{{s}^{-1}}\]
C) \[3.65\times {{10}^{14}}{{s}^{-1}}\]
D) \[5.26\times {{10}^{13}}{{s}^{-1}}\]
Correct Answer: A
Solution :
\[\overline{v}=\frac{1}{\lambda }={{R}_{H}}Z=\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] In Balmer series \[{{n}_{1}}=2\And {{n}_{2}}=3,4,5....\] Last line of the spectrum is called serieslimit Limiting line is the line of shortest wavelength and high energy when \[{{n}_{2}}=\infty \] \[\therefore \]\[\overline{v}=\frac{1}{\lambda }=\frac{{{R}_{H}}}{n_{1}^{2}}=\frac{3.29\times {{10}^{15}}}{{{2}^{2}}}=\frac{3.29\times {{10}^{15}}}{4}\] \[=8.22\times {{10}^{14}}{{s}^{-1}}\]
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