91Ó°ÊÓ

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.

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

$\begin{array}{c}\text{v}=\sqrt{\frac{(\text{9.0}×{\text{10}}^{\text{9}}\frac{\text{N}â‹\dots {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}})×{(\text{1.6}×{\text{10}}^{\text{-19}}\text{ C})}^{\text{2}}}{(\text{9.1}×{\text{10}}^{\text{-31}}\text{kg})×(\text{1.0}×{\text{10}}^{\text{-10}}\text{m})}}\\ =\text{1.6}×{\text{10}}^{\text{6}}\text{m/s}\end{array}$

$\begin{array}{c}{\text{λ}}_{\text{electron}}=\frac{\text{h}}{{\text{p}}_{\text{e}}}=\frac{\text{h}}{{\text{m}}_{\text{e}}\text{v}}\\ {\text{λ}}_{\text{electron}}=\frac{\text{6.63}×{\text{10}}^{\text{-34}}\text{ J}â‹\dots \text{s}}{(\text{9.1}×{\text{10}}^{\text{-31}}\text{kg})×(\text{1.6}×{\text{10}}^{\text{6}}\text{m/s})}\\ {\text{λ}}_{\text{electron}}=\text{4.55}×{\text{10}}^{\text{-10}}\text{m}\end{array}$

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$\left(\text{0.1nm}\right)\text{.}$

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

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