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

The answer is below

Step-by-step explanation:

Using Coulomb's law of electric field which is:

`F=k(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*((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(v^2)/(r) \\\\m = mass\ of \ electron=9.11*10^(-31)g,v=speed\ of\ electron\\\\F=m(v^2)/(r) \\\\v=\sqrt{(F*r)/(m) } \\\\subsituting:\\\\v=\sqrt{(8.22*10^(-8)*5.29*10^(-11))/(9.11*10^(-31)) } \\\\v=2.18*10^6\ m/s\\\\`

`But\ \omega=(v)/(r)=(2.18*10^6)/(5.29*10^(-11)) =4.13*10^(17)\\\\\omega=2\pi f; f=frequency\\\\f=(\omega)/(2\pi) =(4.13*10^(17))/(2\pi) \\\\f=6.57*10^(15)\ Hz`