Difference between revisions of "Classical Bohr Atom"

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[[Radiative Processes in Astrophysics|Course Home]]
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===Short Topical Videos===
 
===Short Topical Videos===
 
* [http://youtu.be/424QV3tD4PE Radiation and Matter, to Order of Magnitude (Aaron Parsons)]
 
* [http://youtu.be/424QV3tD4PE Radiation and Matter, to Order of Magnitude (Aaron Parsons)]
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* [http://en.wikipedia.org/wiki/Diatomic_molecule Diatomic Molecule (Wikipedia)]
 
* [http://en.wikipedia.org/wiki/Diatomic_molecule Diatomic Molecule (Wikipedia)]
 
* [http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_18/node3.html Rotational Levels (Tuckerman, NYU)]
 
* [http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_18/node3.html Rotational Levels (Tuckerman, NYU)]
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===Need to Review?===
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* [[Maxwell Equations]]
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===Related Topics===
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* [[Boltzmann distribution]]
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* [[Atomic and Molecular Quantum Numbers]]
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* [[Rovibrational Transitions]]
  
 
<latex>
 
<latex>
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\begin{itemize}
 
\begin{itemize}
 
\item Electronic Transitions:\par
 
\item Electronic Transitions:\par
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\begin{figure}[h!]
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[[File:Energy_levels_Ryd_alpha_Ly_alpha.png|thumb|400px|Sketch of Rydberg-$\alpha$ and Lyman-$\alpha$ for hydrogen]]
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\end{figure}
 
We'll start with the Bohr atom, and follow a classical derivation/estimation path.  The only quantum mechanics that we need to  
 
We'll start with the Bohr atom, and follow a classical derivation/estimation path.  The only quantum mechanics that we need to  
inject is that angular momentum comes in units of $\hbar$.  So let's begin by quantizing the angular momentum associated with the electron's orbit around the nucleus.  We'll define the radius of the electron's orbit as $a_0$, and the velocity that the electron travels at will be $v_e$.  This gives us the following expression for the angular momentum for $n$ units of $\hbar$:
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inject is that angular momentum comes in units of $\hbar$.  So let's begin by quantizing the angular momentum associated with the electron's orbit around the nucleus.  We'll define the radius of the electron's orbit as $a_0$ (the Bohr radius), and the velocity that the electron travels at will be $v_e$.  This gives us the following expression for the electron's angular momentum for $n$ units of $\hbar$:
$$m_ev_ea_0=n\hbar$$
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\begin{equation}
If we balance the force required to keep the $e^-$ in a circular orbit
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m_ev_ea_0=n\hbar
with the electric force:
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\end{equation}
$${m_ev_e^2\over a_0}={Ze^2\over a_0^2}$$
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Next, we balance the force required to keep the $e^-$ in a circular orbit
$$a_0={\hbar^2\over m_ee^2Z}\approx0.52{\AA\over Z}$$
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against the electric force:
 +
\begin{equation}
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m_e\frac{v_e^2}{a_0}=\frac{Ze^2}{a_0^2}
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\end{equation}
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This gives us the following solution for the Bohr radius:
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\begin{equation}
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a_0=\frac{\hbar^2}{m_eZe^2}\approx0.52\frac{\AA}{Z}
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\end{equation}
 
A {\it Rydberg} is the energy required to ionize an H atom from the ground
 
A {\it Rydberg} is the energy required to ionize an H atom from the ground
state.  It is $\sim13.6eV$.  We can estimate it by integrating the electric
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state.  It is $\sim13.6eV$.  We can estimate it as half of the energy we get integrating the electric
force from $r=a_0$ to $r=\infty$, but in reality, there is another factor of
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force from $r=a_0\rightarrow\infty$, (the other half is the kinetic energy of the orbiting electron):
2:
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\begin{equation}
$$\boxed{Rydberg={Ze^2\over2a_0}={Z^2e^4m_e\over2\hbar^2}=13.6\cdot Z^2eV}$$
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{\rm Rydberg}\equiv\frac{Ze^2}{2a_0}=\frac{Z^2e^4m_e}{\hbar^2}=13.6\cdot Z^2eV
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\end{equation}
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\item Fine Structure:\par
 
\item Fine Structure:\par
 
Fine Structure comes from the interaction of the magnetic moment of the $e^-$
 
Fine Structure comes from the interaction of the magnetic moment of the $e^-$
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energy of a dipole interaction is $E=\vec\mu\cdot\bfield$, so we'd expect
 
energy of a dipole interaction is $E=\vec\mu\cdot\bfield$, so we'd expect
 
that:
 
that:
$$\Delta E\sim\mu B$$
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\begin{equation}
To estimate B: $B\sim{Ze\over a_0^2}{v\over c}$, and ${v\over c}\sim
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\Delta E\sim\mu B
{e^2\over\hbar c}=\alpha=\inv{137}$ ($\alpha$ is the fine structure constant):
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\end{equation}
$$B\sim{Ze^3\over\hbar ca_0^2}$$
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We can estimate B by just applying a Lorentz boost to the electric field of the nucleus:
Estimating $\mu$: The $\bfield$ of a dipole goes as $B_{di}\sim{\mu\over r^3}$,
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\begin{equation}
so $\mu\sim B_e\eval{r_e}r_e^3$, where $r_e$ is the classical $e^-$ radius.
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B\sim\frac{Ze}{a_0^2}\frac{v}{c},
Estimating that the rest mass energy of the $e^-$ should be about the  
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\end{equation}
electrostatic potential energy left in the $e^-$, $m_ec^2\sim{e\over r_e}$:
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with $\frac{v}{c}\sim\frac{Ze^2}{\hbar c}$.  For reference, $\frac{v}{c}$ for $Z=1$ evaluates to $\frac{e^2}{\hbar c}\equiv\alpha\approx\inv{137}$, which is the fine structure constant
$$r_e\sim{e^2\over m_ec^2}$$
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We can estimate $B_e$ by reverting to Maxwell's equation:
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Next we need to estimate the magnetic dipole of an electron $\mu_e$. The $\bfield$ of a magnetic dipole goes as $B_{di}\sim\frac{\mu}{r^3}$,
$$\dcb={4\pi J\over c}$$
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so we can estimate the magnetic dipole of an electron as something that produces the $\bfield$ of a spinning electron at $r_e$, the classical
$${B_e\over2\pi r_e}={4\pi\over c}{I\over4\pi r_e^2}$$
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electron radius:
Again, we estimate that $I\sim{e\over t_{spin}}$, and because the electron
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\begin{equation}
spin is quantized, $\hbar=m_er_e^2{2\pi\over t_{spin}}$.
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\mu\sim B_e\eval{r_e}r_e^3.
 +
\end{equation}
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We can estimate $r_e$ by setting the rest mass energy of the $e^-$ to the  
 +
electrostatic potential energy of the electron:
 +
\begin{align}
 +
m_ec^2&\sim\frac{e^2}{r_e}\\
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r_e&\sim\frac{e^2}{m_ec^2}\\
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\end{align}
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To estimate $B_e$, we can use the [[Maxwell Equations]]:
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\begin{align}
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\dcb&=\frac{4\pi}{c} J\\
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\frac{B_e}{2\pi r_e}&=\frac{4\pi}{c}\frac{I}{4\pi r_e^2}\\
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\end{align}
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We can estimate the current, $I$ from the spin timescale of the electron, $I\sim\frac{e}{t_{spin}}$, and because the electron
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spin is quantized, $\hbar=m_er_e^2\frac{2\pi}{t_{spin}}$.
 
After some algebra, we get that:
 
After some algebra, we get that:
$$\mu_e={e\hbar\over2m_ec}$$
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\begin{equation}
 +
\mu_e=\frac{e\hbar}{2m_ec}
 +
\end{equation}
 
\centerline{(Bohr magneton for $e^-$)}
 
\centerline{(Bohr magneton for $e^-$)}
$$\mu_p={Ze\hbar\over2m_pc}$$
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\begin{equation}
 +
\mu_p=\frac{Ze\hbar}{2m_pc}
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\end{equation}
 
\centerline{(Bohr magneton for nucleus)}
 
\centerline{(Bohr magneton for nucleus)}
Getting back to the energy:
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Finally, we plug these in for our expression for the energy:
$$\begin{aligned}\Delta E&\sim{e\hbar\over2m_ec}{Ze\over a_0^2}\alpha\\  
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\begin{align}
&\sim Z^4\alpha^2\cdot Ryd\\ \end{aligned}$$
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\Delta E&\sim\frac{e\hbar}{2m_ec}\frac{Ze}{ a_0^2}Z\alpha\\  
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&\sim Z^4\alpha\cdot {\rm Ryd}\\
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\end{align}
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where we take the Rydberg to be 13.6 eV and have factored out its $Z$ dependence explicitly.
  
 
\item Hyperfine Structure:\par
 
\item Hyperfine Structure:\par

Latest revision as of 16:59, 5 December 2017

Course Home

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Reference Material[edit]

Need to Review?[edit]

Related Topics[edit]

Order of Magnitude Interaction of Radiation and Matter

It turns out that you can tell a cute little story about the energy levels in hydrogen-like atoms using a simple Bohr model of the hydrogen atom and classical physics. You just need a couple small doses of quantum mechanics in the beginning to get you started on the right track.

Energy Levels

  • Electronic Transitions:
Sketch of Rydberg- and Lyman- for hydrogen

We’ll start with the Bohr atom, and follow a classical derivation/estimation path. The only quantum mechanics that we need to inject is that angular momentum comes in units of . So let’s begin by quantizing the angular momentum associated with the electron’s orbit around the nucleus. We’ll define the radius of the electron’s orbit as (the Bohr radius), and the velocity that the electron travels at will be . This gives us the following expression for the electron’s angular momentum for units of :

Next, we balance the force required to keep the in a circular orbit against the electric force:

This gives us the following solution for the Bohr radius:

A Rydberg is the energy required to ionize an H atom from the ground state. It is . We can estimate it as half of the energy we get integrating the electric force from , (the other half is the kinetic energy of the orbiting electron):

  • Fine Structure:

Fine Structure comes from the interaction of the magnetic moment of the with the a caused by the Lorentz-transformed Coulomb field of the proton (generated by the ’s motion). The energy of a dipole interaction is , so we’d expect that:

We can estimate B by just applying a Lorentz boost to the electric field of the nucleus:

with . For reference, for evaluates to , which is the fine structure constant.

Next we need to estimate the magnetic dipole of an electron . The of a magnetic dipole goes as , so we can estimate the magnetic dipole of an electron as something that produces the of a spinning electron at , the classical electron radius:

We can estimate by setting the rest mass energy of the to the electrostatic potential energy of the electron:

To estimate , we can use the Maxwell Equations:

We can estimate the current, from the spin timescale of the electron, , and because the electron spin is quantized, . After some algebra, we get that:

(Bohr magneton for )

(Bohr magneton for nucleus) Finally, we plug these in for our expression for the energy:

where we take the Rydberg to be 13.6 eV and have factored out its dependence explicitly.

  • Hyperfine Structure:

Instead of using the Bohr magneton, we use the intrinsic magnetic moment (spin) of the nucleus:

Thus:

Note that Fine and Hyperfine are magnetic dipole transitions. level transitions are electric dipole transitions. Magnetic dipole transitions are generally weaker.

  • Vibrational Transitions in Molecules:

Our general technique with vibrational transitions is to model them as harmonic oscillators. Thus, they should have the characteristic harmonic energy series:

For a harmonic oscillator, . We estimate that since the force for a spring is , and that force should be about the Coulomb force on ’s. If we say that atoms stretch with respect to each other about a Bohr radius:

where A is the atomic mass # of our atoms.

  • Rotational Transitions in Molecules:

The thing to remember is that angular momentum comes in units of .