A) \[1.8\times {{10}^{15}}Hz\]
B) \[3.2\times {{10}^{15}}Hz\]
C) \[4.7\times {{10}^{5}}Hz\]
D) \[6.9\times {{10}^{15}}Hz\]
Correct Answer: B
Solution :
According to Bohr?s theory the wavelength of the radiation \[\frac{1}{\lambda }={{Z}^{2}}R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\frac{v}{c}={{Z}^{2}}R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[v=c{{Z}^{2}}R\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[2.7\times {{10}^{15}}=c{{Z}^{2}}R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{2}^{2}}} \right]\] \[v=c{{Z}^{2}}R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{3}^{2}}} \right]\] \[\frac{v}{2.7\times {{10}^{-16}}}=\frac{1-\frac{1}{9}}{1-\frac{1}{4}}=\frac{8}{9}\times \frac{4}{3}=\frac{32}{27}\] or frequency, \[v=\frac{32}{27}\times 2.7\times {{10}^{15}}Hz\] \[=3.2\times {{10}^{15}}Hz\]
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