A) \[4.2\times {{10}^{15}}Hz\]
B) \[4.2\text{ }J/{{g}^{o}}C\]
C) \[3\times {{10}^{9}}Hz\]
D) \[{{20}^{o}}C\]
Correct Answer: C
Solution :
We know that \[\frac{m{{v}^{2}}}{r}={{\left( \frac{2{{e}^{2}}}{4\pi {{\varepsilon }_{0}}mr} \right)}^{1/2}}\] Also frequency \[f=\frac{v}{2\pi r}={{\left( \frac{2{{e}^{2}}}{4\pi {{\varepsilon }_{0}}mv} \right)}^{1/2}}\frac{1}{2\pi r}\] \[f=\frac{(1.414)\,{{(9\times {{10}^{9}})}^{1/2}}1.6\times {{10}^{-19}}}{{{(9.1\times {{10}^{-31}})}^{1/2}}\times 2\pi {{({{10}^{-10}})}^{3/2}}}\] \[=3.6\times {{10}^{15}}Hz\]
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