91Ó°ÊÓ

The Coulomb force between the electron and the proton provides the

centripetal force that keeps the electron in circular orbit about the proton:

$\frac{\mathbf{k}{\mathbf{\left|}\mathbf{e}\mathbf{\right|}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{e}}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

The smallest orbital radius is$r_1=a_0=52.9×{10}^{−12}\text{″¾}$. The corresponding speed of the electron is obtained as follows:

$\begin{array}{c}{v}_1=\sqrt{\frac{k{\left|e\right|}^2}{m_e{r}_1}}\\ =\sqrt{\frac{k{\left|e\right|}^2}{m_e{a}_0}}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}{v}_1=\sqrt{\frac{(9×{10}^9\text{ }\frac{{\text{Nm}}^{\text{2}}}{{\text{C}}^{\text{2}}}){(1.60×{10}^{−19}\text{ C})}^2}{(9.11×{10}^{−31}\text{ k²µ})(52.9×{10}^{−12}\text{″¾})}}\\ =2.19×{10}^6\text{ }\frac{\text{m}}{\text{s}}\end{array}$

The radius of the second smallest orbit is$r_2={\left(2\right)}^2{a}_0=4a_0$.

Thus, the speed of the electron is as follows:

$\begin{array}{c}{v}_2=\sqrt{\frac{k{\left|e\right|}^2}{m_e{r}_2}}\\ =\sqrt{\frac{k{\left|e\right|}^2}{m_e\left(4a_0\right)}}\end{array}$

Substitute the values and solve as:

$\begin{array}{c}{v}_2=\frac{1}{2}{v}_1\\ =\frac{1}{2}(2.19×{10}^6\text{ }\frac{\text{m}}{\text{s}})\\ =1.09×{10}^6\text{ }\frac{\text{m}}{\text{s}}\end{array}$

Since the speed is inversely proportional to$r^{\frac{1}{2}}$ , the speed of the electron will decrease if it moves to larger orbits