Question

In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular motion about a proton with the centripetal force caused by Coulomb attraction. He predicted the radius of the electron's orbit to be 5.29 ✕ 10−11 m. Calculate the speed of the electron and the frequency of its circular motion.

Answer 1

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

$F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s$

$But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}} =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz$