In general, we might say that the wavelengths allowed a bound particle are those of a typical standing wave,$\mathit{\lambda }\mathbf{=}\mathbf{2}\mathit{L}\mathbf{/}\mathit{n}$ , where is the length of its home. Given that $\mathit{\lambda }\mathbf{=}\mathit{h}\mathbf{/}\mathit{p}$, we would have $\mathit{p}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathit{h}\mathbf{/}\mathbf{2}\mathit{L}$, and the kinetic energy, ${\mathit{p}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathit{m}$, would thus be ${\mathit{n}}^{\mathbf{2}}{\mathit{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}\mathit{m}{\mathit{L}}^{\mathbf{2}}$. These are actually the correct infinite well energies, for the argumentis perfectly valid when the potential energy is 0 (inside the well) and is strictly constant. But it is a pretty good guide to how the energies should go in other cases. The length allowed the wave should be roughly the region classically allowed to the particle, which depends on the “height” of the total energy E relative to the potential energy (cf. Figure 4). The “wall” is the classical turning point, where there is nokinetic energy left: $\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathit{U}$. Treating it as essentially a one-dimensional (radial) problem, apply these arguments to the hydrogen atom potential energy (10). Find the location of the classical turning point in terms of E , use twice this distance for (the electron can be on both on sides of the origin), and from this obtain an expression for the expected average kinetic energies in terms of E . For the average potential, use its value at half the distance from the origin to the turning point, again in terms of . Then write out the expected average total energy and solve for E . What do you obtain

for the quantized energies?

Step by step solution

01

Given data

The momentum of a particle classically bound in a region of length L is

$p=\frac{nh}{2L}$ ..... (I)

Here h is the Planck's constant.

02

Hydrogen atom potential energy

The potential energy of a Hydrogen atom of radius r is

$U=-\frac{e^2}{4\pi {\epsilon }_{0}r}$ ..... (II)

Here e is the magnitude of charge of an electron and ${\epsilon }_0$ is the permittivity of free space.

03

Step 3:Determining the total energy of a particle stuck inside a Hydrogen atom radius

From equation (II) the radius at which the kinetic energy is zero is

$\begin{array}{rcl}E& =& \frac{e^2}{4\pi {\epsilon }_0{r}_{\max }}\\ r_{\max }& =& \frac{e^2}{4\pi {\epsilon }_{0}E}\\ & & \end{array}$

The classically allowed region is twice this value, that is

$\begin{array}{rcl}L& =& 2r_{\max }\\ & =& \frac{2e^2}{4\pi {\epsilon }_{0}E}\\ & =& \frac{e^2}{2\pi {\epsilon }_{0}E}\\ & & \end{array}$

From equation (I), the momentum is

$\begin{array}{rcl}p& =& \frac{nh}{2\times \frac{e^2}{2\pi {\epsilon }_{0}E}}\\ & =& \frac{nh\pi {\epsilon }_{0}E}{e^2}\\ & & \end{array}$

The kinetic energy is

$\begin{array}{rcl}T& =& \frac{p^2}{2m}\\ & =& \frac{{\left(\frac{nh\pi {\epsilon }_{0}E}{e^2}\right)}^2}{2m}\\ & =& \frac{n^2{h}^2{\pi }^2{\epsilon }_0^2{E}^2}{2m{e}^4}\\ & & \end{array}$

From equation (II), the potential energy at half the classical limit is

$\begin{array}{rcl}U\left(r=\frac{r_{\max }}{2}\right)& =& -\frac{e^2}{4\pi {\epsilon }_0\frac{r_{\max }}{2}}\\ & =& -\frac{e^2}{4\pi {\epsilon }_0\times \frac{1}{2}\times \frac{e^2}{4\pi {\epsilon }_{0}E}}\\ & =& -2E\\ & & \end{array}$

The total energy is then

$\begin{array}{rcl}E& =& T+U\\ & =& \frac{n^2{h}^2{\pi }^2{\epsilon }_0^2{E}^2}{2m{e}^4}-2E\\ 3E& =& \frac{n^2{h}^2{\pi }^2{\epsilon }_0^2{E}^2}{2m{e}^4}\\ E& =& \frac{6m{e}^4}{n^2{h}^2{\pi }^2{\epsilon }_0^2}\\ & & \end{array}$

Thus the quantized energies are $E=\frac{6m{e}^4}{n^2{h}^2{\pi }^2{\epsilon }_0^2}$.

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