Question 42.2: Orbiting Electrons in Hydrogen Let’s examine Rutherford’s mo......
Orbiting Electrons in Hydrogen
Let’s examine Rutherford’s model more closely. Consider hydrogen, the simplest atom, with a nucleus consisting of a single proton that is orbited by a single electron. Model the nucleus as a sphere with a radius of roughly 10^{-15} m, and assume the electron orbits the center of the nucleus in a circle with a radius of roughly 10^{-10} m. Find the magnitude of the force exerted on the electron by the nucleus and the electron’s speed.
Interpret and Anticipate
We could have done this problem in Chapter 23. Coulomb’s law gives the force exerted on the electron, and because no other forces are present, this is the centripetal force. We can find the electron’s speed from its centripetal acceleration.
Solve
Use Coulomb’s law (Eq. 23.3) to find the force on the electron. As usual, quantity r is the distance from the center of the sphere to the particle; in this case r is the radius of the electron’s orbit. For hydrogen, the nucleus and the electron have the same magnitude of charge, e.
This is the centripetal force, so the electron’s speed is found using F_c = m\nu^2/r (Eq. 6.7).
\begin{aligned}& F_c=F_E=m_e \frac{\nu^2}{r} \\& \nu=\sqrt{\frac{r F_E}{m_e}}=\sqrt{\frac{\left(10^{-10} m \right)\left(2.3 \times 10^{-8} N \right)}{9.11 \times 10^{-31} kg }} \\& \nu=1.6 \times 10^6 m / s\end{aligned}Check and Think
The electron’s orbital speed is about 0.5% the speed of light, so there is no need to use special relativity. Our estimate of this speed seems reasonable (not too fast, not too slow), which partly explains why the solar system model of the atom along with classical physics explains so many phenomena. However, Rutherford’s solar system model cannot account for the hydrogen spectrum (Fig. 42.5), the subject of part 2 of our CASE STUDY .
