Question

If an electron is moving around a nucleus of charge 2e in a circular orbit of radius $ 3.6\times {{10}^{15}}Hz $ then calculate the initial frequency of light emitted by the electron

• 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 $

Answer 1

Answer: (C)

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 $