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. $4.2J/g^{o}C$
• C. $3\times {10}^{9}Hz$
• D. ${20}^{o}C$

Answer 1

Answer: (C)

We know that $\frac{m{v}^2}{r}={\left(\frac{2e^2}{4\pi {\epsilon }_{0}mr}\right)}^{1/2}$ Also frequency $f=\frac{v}{2\pi r}={\left(\frac{2e^2}{4\pi {\epsilon }_{0}mv}\right)}^{1/2}\frac{1}{2\pi r}$ $f=\frac{(1.414){(9\times {10}^9)}^{1/2}1.6\times {10}^{-19}}{{(9.1\times {10}^{-31})}^{1/2}\times 2\pi {({10}^{-10})}^{3/2}}$ $=3.6\times {10}^{15}Hz$