The Hydrogen Atom¶

https://www.youtube.com/watch?v=-Y0XL-K0jy0

Quiz¶

https://www.youtube.com/watch?v=9GOuZAh9Wg0

Mystery of the Hydrogen Atom¶

Because the proton’s mass is much greater than the electron’s mass, we shall assume that the proton is fixed in place. So, the atom is a fixed potential trap with the electron moving around inside it.

A hydrogen atom contains an electron that is trapped by the Coulomb force it experiences from the proton, which is the nucleus of the atom.

Under Newtonian laws, the electron would move around the proton, like planets around the Sun, i.e. $\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}=m\frac{v^2}{r}$

Multiplying by −r, we obtain $E_c=-\frac{e^2}{4\pi \epsilon _0 r}=-mv^2=-2E_k$

Alternatively, the total energy of the electron is$E=E_k+E_c=\frac{E_c}{2}=-E_k$

However, any charged particle which moves in a curved path will emit electromagnetic radiation, hence losing energy continuously. Why doesn’t the electrical attraction between the electron and the positive charge simply cause the two to collapse together?

One clue lies in the experimental fact that a hydrogen atom can emit and absorb only four wavelengths in the visible spectrum (656 nm, 486 nm, 434 nm, and 410 nm).

The Bohr Model of Hydrogen (1913)¶

Bohr made two bold (and completely unjustified) assumptions:

The electron in a hydrogen atom orbits the nucleus in a circlemuch like Earth orbits the Sun.
The magnitude of the angular momentum $\vec{L}$ of the electron in its orbit is restricted (quantized) to the values $L = n\bar{h}$, for $n = 1,2,3,....$

However, as successful as his theory was on the four visible wavelengths and on why the atom did not simply collapse, it turned out to be quite wrong in almost every other aspect of the atom.

Physical meaning of the assumption¶

The de Broglie wavelength $λ$ of a free particle with momentum p is $λ = h/p = h/mv$.For an electron in a hydrogen atom whose orbital radius $r$, the above equation leads to $λ/r = h/rmv = h/L$. Therefore, we obtain $2πr/λ = L/\bar{h}$.

$L=n\bar{h}$ means that the length of the orbit is an integer multiple of $λ$. Namely, the phase of the electron wave function returns to the initial value by moving for one cycle of the orbit.

Analysis based on Bohr’s model¶

Bohr to quantize the electron orbit $L=rmv=n\bar{h} \Rightarrow v=\frac{n\bar{h}}{mr}$
Combining with the Newtonian result $\frac{e^2}{4\pi\epsilon_0r^2}=m\frac{v^2}{r}$ we find $r_n =n^2a_B$. where the characteristic length $a_B=\frac{\bar{h}^2}{me^2/(4\pi\epsilon_0)}=0.529 Å$

In the Bohr model of the hydrogen atom, the electron’s orbital radius r is quantized and the smallest possible orbital radius (for $n = 1$) is $a_B$, which is called the Bohr radius.

Can you obtain the length scale, alternatively, from dimension analysis?sIn order that the attraction between electron and nucleus does not simply collapse them together in the Bohr model, the electron should not get any closer to the nucleus than orbital radius aB by taking
The energy of the hydrogen atom, according to the Bohr model, is then $E_n=\frac{1}{2}mv^2-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}=-\frac{E_R}{n^2}$ where $E_R=\frac{me^4/(4\pi\epsilon_0)^2}{2\bar{h}^2}=13.6eV$
We still have, for each orbit $E=E_k+E_c=\frac{E_c}{2}=-E_k$

The Hydrogen Spectrum¶

The energy of a hydrogen atom (or, equivalently, of its electron) changes when the atom emits or absorbs light. Emission and absorption involve a quantum of light according to $\bar{h}ω_{nm} =E_R(\frac{1}{n^2} −\frac{1}{m^2})$ for integers $m > n$.

The wavelengths of the emitted or absorbed light are given by $ \frac{1}{\lambda}=\frac{E_R}{hc}(\frac{1}{n^2}-\frac{1}{m^2})$

$1\ Ångström - 1×10^{−10}m$

Combinations of Physical Constants¶

GS Energy from Uncertainty Principle[Ground State]¶

The ground-state energy is the lowest energy allowed by Heisenberg’s uncertainty principle.

For a hydrogen atom, the size of the wave function, $∆r$, is the uncertainty in position.
The uncertainty in momentum is, roughly speaking, $∆p ∼ \bar{h}/∆r$, by the uncertainty principle.
The energy of the electron can be estimated to be $E ∼ \frac{(\triangle p)^2}{2m}-\frac{e^2}{4\pi\epsilon_0\triangle r}=\frac{\bar{h}^2}{2m(\triangle r)^2}-\frac{e^2}{4\pi\epsilon_0\triangle r}$$(Mean(p)=0\ Thus\ use\ \triangle p\ to\ approximate\ p)$

To find the minimal energy, we solve for $\triangle r$ $\Rightarrow\frac{dE}{d(\triangle r)}=0 \Rightarrow \triangle r = \frac{\bar{h}^2}{me^2/(4\pi\epsilon_0)}=a_B$ and $E=-\frac{me^4/(4\pi\epsilon_0)^2}{2\bar{h}^2}=-E_R$

The energy of the ground state (or any stationary state) is uniquely determined. 「Though the uncertainty of $t\ \&\ x$」

This is because of the energy-time uncertainty principle, $∆t · ∆E ≥ \bar{h}/2$.

In the extreme case of a stationary state, $∆t = ∞$, so we have $∆E = 0$.

Note, however, both kinetic energy and potential energy have uncertainties, due to the uncertainties of position and momentum.

Hydrogen is a three-dimensional, finite electron trap, with walls that vary in depth with distance.

Schroedinger's Equation for the H-atom¶

Derive¶

Ground State Wave Function¶

$ψ_{100}(\vec{r})=R_{10}(r)=\frac{1}{\sqrt{\pi}a_B^{3/2}}e^{\frac{r}{a_B}}$

Note that the hydrogen atom in its ground state has zero angular momentum ($l = 0$ more details will be discussed in the next lecture), which is not predicted in the Bohr model.)

The probability that the electron can be detected in any given (infinitesimal) volume element $dV$ located at radius $r$ from the center of the atom is $|ψ_{100}(\vec{r})|^2dV.$

Define radial Probability density $P(r)$ such that $P(r)dr=|ψ_{100}(\vec{r})|^2dV\ (dV=4\pi r^2)$
$P(r)$ takes a maximum at $r=a_B$
All we can ever know abount the location of the electron in the ground state of the hydrogen atom is the radial probability density
Dot Plot

Excited States of the Hydrogen Atom¶

Bohr’s Correspondence Principle¶

最后更新: 2024年3月25日 12:53:47
创建日期: 2023年12月26日 15:12:27