$\begin{array}{1 1}(a)\;7.29\times 10^7cmsec^{-1}&(b)\;7.30\times 10^2cmsec^{-1}\\(c)\;8.39\times 10^7cmsec^{-1}&(d)\;7.29\times 10^3cmsec^{-1}\end{array}$

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$\mu_n=\sqrt{\big(\large\frac{ze^2}{mr_n}\big)}$

Radius of $III$ orbit =$r_1\times n^2=0.529\times 10^{-8}\times 9cm$

$\mu_n=\sqrt{\big(\large\frac{1\times (4.083\times 10^{-10})^2}{9.108\times 10^{-28}\times 0.529\times 10^{-8}\times 9}\big)}$

$\mu_n=7.29\times 10^7cmsec^{-1}$

Hence (a) is the correct answer.

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