Question 21.73: In an early model of the hydrogen atom (the Bohr model), the......
In an early model of the hydrogen atom (the Bohr model), the electron orbits the proton in uniformly circular motion. The radius of the circle is restricted (quantized) to certain values given by
r = n²a_0, for n = 1, 2, 3, . . . ,
where a_0 = 52.92 pm. What is the speed of the electron if it orbits in (a) the smallest allowed orbit and (b) the second smallest orbit? (c) If the electron moves to larger orbits, does its speed increase, decrease, or stay the same?
(a) The Coulomb force between the electron and the proton provides the centripetal force that keeps the electron in circular orbit about the proton:
\frac{k|e|^2}{r^2}=\frac{m_e ν^2}{r}
The smallest orbital radius is r_1=a_0=52.9 \times 10^{-12} \,m . The corresponding speed of the electron is
\begin{aligned} ν_1 & =\sqrt{\frac{k|e|^2}{m_e r_1}}=\sqrt{\frac{k|e|^2}{m_e a_0}}=\sqrt{\frac{\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right)\left(1.6 \times 10^{-19} \,C \right)^2}{\left(9.11 \times 10^{-31} \,kg \right)\left(52.9 \times 10^{-12} \,m \right)}} \\ & =2.19 \times 10^6 \,m / s . \end{aligned}
(b) The radius of the second smallest orbit is r_2=(2)^2 a_0=4 a_0 . Thus, the speed of the electron is
\begin{aligned} ν_2 & =\sqrt{\frac{k|e|^2}{m_e r_2}}=\sqrt{\frac{k|e|^2}{m_e\left(4 a_0\right)}}=\frac{1}{2} ν_1=\frac{1}{2}\left(2.19 \times 10^6 \,m / s \right) \\ & =1.09 \times 10^6 \,m / s . \end{aligned}
(c) Since the speed is inversely proportional to r^{1/ 2} , the speed of the electron will decrease if it moves to larger orbits.