Question

In the hydrogen atom, the electron’s orbit, not necessarily circular, extends to a distance of a about an angstrom $\mathbf{(}\mathbf{\text{1Å}}\mathbf{=}\mathbf{0}\mathbf{.1}\mathbf{}\mathbf{\text{\hspace{0.17em}nm}}\mathbf{)}$from the proton. If it is to move about as a compact classical particle in the region where it is confined, the electron’s wavelength had better always be much smaller than an angstrom. Here we investigate how large might be the electron’s wavelength. If orbiting as a particle, its speed at $\mathbf{\text{1Å}}$could be no faster than that for circular orbit at that radius. Why? Find the corresponding wavelength and compare it to$\mathbf{\text{1Å}}$ . Can the atom be treated classically?

Short answer

The electron is not a classical particle and cannot be regarded as such.

Step-by-step solution

Given.

No longer working with circular orbits if the electrons' speed around the nucleus reaches one angstrom. Electrons must maintain the same radius; otherwise, their velocity will be dependent on their position, and will not be able to link a particular wavelength with this classical particle. Now, assuming the electron is a classical particle orbiting the nucleus in a circular orbit, may apply Newton's second law to a circular motion, with the electric force as the dominating force.

$\begin{array}{c}{\text{F}}_{\text{E}}=\frac{{\text{mv}}^{\text{2}}}{\text{r}}\\ \frac{{\text{ke}}^{\text{2}}}{{\text{r}}^{\text{2}}}=\frac{{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}}{\text{r}}\\ \text{v}=\sqrt{\frac{{\text{ke}}^{\text{2}}}{{\text{m}}_{\text{e}}\text{r}}}\end{array}$

Substituting formula.

$\begin{array}{c}\text{v}=\sqrt{\frac{(\text{9.0}\times {\text{10}}^{\text{9}}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}})\times {(\text{1.6}\times {\text{10}}^{\text{-19}}\text{\hspace{0.17em}C})}^{\text{2}}}{(\text{9.1}\times {\text{10}}^{\text{-31}}\text{kg})\times (\text{1.0}\times {\text{10}}^{\text{-10}}\text{m})}}\\ =\text{1.6}\times {\text{10}}^{\text{6}}\text{m/s}\end{array}$

$\begin{array}{c}{\text{λ}}_{\text{electron}}=\frac{\text{h}}{{\text{p}}_{\text{e}}}=\frac{\text{h}}{{\text{m}}_{\text{e}}\text{v}}\\ {\text{λ}}_{\text{electron}}=\frac{\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.17em}J}\cdot \text{s}}{(\text{9.1}\times {\text{10}}^{\text{-31}}\text{kg})\times (\text{1.6}\times {\text{10}}^{\text{6}}\text{m/s})}\\ {\text{λ}}_{\text{electron}}=\text{4.55}\times {\text{10}}^{\text{-10}}\text{m}\end{array}$

dimension in the problem to be treated as a classical particle, however, this is not the case here.

We can't treat the electron as a classical particle because its wavelength$\left(\text{0.455nm}\right)$ is bigger than its orbital radius$\left(\text{0.1nm}\right)\text{.}$

${\text{λ}}_{\text{electron}}=\text{4.55}\times {\text{10}}^{\text{-10}}\text{m}$

The electron is not a classical particle and cannot be regarded as such.