Question

In a classic model of the hydrogen atom, the electron moves around the proton in a circular orbit of radius$0.053nm.$

a. What is the electron's orbital frequency?

b. What is the effective current of the electron?

Short answer

a. The electron's orbital frequency is$6.56\times {10}^{15}\mathrm{Hz}$

b. The effective current of the electron is $1.05\mathrm{mA}.$

Step-by-step solution

Given Information (Part a)

Hydrogen atom circular orbit of the radius is$0.053nm$.

Explanation (Part a)

The centripetal force is the electrostatic one, then

$\frac{k{e}^2}{R^2}=F=m{\omega }^{2}R$

The angular frequency is just $\omega =2\pi f$,

$\frac{k{e}^2}{R^2}=F=4{\pi }^{2}m{f}^{2}R$

Rearrange for the frequency as,

$f^2=\frac{k{e}^2}{4{\pi }^{2}m{R}^3},$

The frequency will be given by

$f=\frac{e}{2\pi }\sqrt{\frac{k}{m{R}^3}}$

Substituting the values,

$f=\left(\frac{{\displaystyle 1.6\cdot {10}^{-19}C}}{{\displaystyle 2\pi }}\sqrt{\frac{{\displaystyle 9\times {10}^9}}{{\displaystyle 9.1\times {10}^{-31}kg\cdot {\left(5.3\cdot {10}^{-11}m\right)}^3}}}\right)$

$=6.56\cdot {10}^{15}\mathrm{Hz}$

Final Answer (Part a)

Hence, the hydrogen atom circular orbit of the radius is$6.56\times {10}^{15}\mathrm{Hz}$.

Given Information (Part b) 

Hydrogen atom circular orbit of the radius is$0.053nm.$

Explanation (Part b) 

In one second, the effective current is equal to the frequency times the elementary charge, and this is the total charge that is passed.

That is,

$q_{eff}=fe=\left(6.56\times {10}^{15}Hz\times 1.6\times {10}^{-19}C\right)$

$q_{eff}=1.05\times {10}^{-3}\mathrm{A}.$

Final Answer (Part b) 

The effective current of the electron is$1.0mA.$