# Difference between revisions of "Classical Bohr Atom"

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\end{figure} | \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 $\hbar$: | + | 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 $\hbar$: |

− | + | \begin{equation} | |

− | + | m_ev_ea_0=n\hbar | |

− | + | \end{equation} | |

− | + | Next, we balance the force required to keep the $e^-$ in a circular orbit | |

− | + | against the electric force: | |

+ | \begin{equation} | ||

+ | m_e\frac{v_e^2}{a_0}=\frac{Ze^2}{a_0^2} | ||

+ | \end{equation} | ||

+ | This gives us the following solution for the Bohr radius: | ||

+ | \begin{equation} | ||

+ | a_0=\frac{\hbar^2}{m_eZe^2}\approx0.52\frac{\AA}{Z} | ||

+ | \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 | + | state. It is $\sim13.6eV$. We can estimate it as half of the energy we get integrating the electric |

− | force from $r=a_0 | + | force from $r=a_0\rightarrow\infty$, (the other half is the kinetic energy of the orbiting electron): |

− | + | \begin{equation} | |

− | + | {\rm Rydberg}\equiv\frac{Ze^2}{2a_0}=\frac{Z^2e^4m_e}{\hbar^2}=13.6\cdot Z^2eV | |

+ | \end{equation} | ||

+ | |||

\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: | ||

− | + | \begin{equation} | |

− | + | \Delta E\sim\mu B | |

− | {e^2 | + | \end{equation} |

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

− | + | \begin{equation} | |

− | so $\mu\sim B_e\eval{r_e}r_e^3 | + | B\sim\frac{Ze}{a_0^2}\frac{v}{c}, |

− | + | \end{equation} | |

− | electrostatic potential energy | + | 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. |

− | + | ||

− | + | Next we need to estimate the magnetic dipole of an electron $\mu_e$. The $\bfield$ of a magnetic dipole goes as $B_{di}\sim\frac{\mu}{r^3}$, | |

− | + | 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 | |

− | + | electron radius: | |

− | + | \begin{equation} | |

− | spin is quantized, $\hbar=m_er_e^2{2\pi | + | \mu\sim B_e\eval{r_e}r_e^3. |

+ | \end{equation} | ||

+ | 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}\\ | ||

+ | r_e&\sim\frac{e^2}{m_ec^2}\\ | ||

+ | \end{align} | ||

+ | To estimate $B_e$, we can use Maxwell's equation: | ||

+ | \begin{align} | ||

+ | \dcb&=\frac{4\pi}{c} J\\ | ||

+ | \frac{B_e}{2\pi r_e}&=\frac{4\pi}{c}\frac{I}{4\pi r_e^2}\\ | ||

+ | \end{align} | ||

+ | We can estimate the current, $I$ from the spin timescale of the electron, $I\sim\frac{e}{t_{spin}}$, and because the electron | ||

+ | 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: | ||

− | + | \begin{equation} | |

+ | \mu_e=\frac{e\hbar}{2m_ec} | ||

+ | \end{equation} | ||

\centerline{(Bohr magneton for $e^-$)} | \centerline{(Bohr magneton for $e^-$)} | ||

− | + | \begin{equation} | |

+ | \mu_p=\frac{Ze\hbar}{2m_pc} | ||

+ | \end{equation} | ||

\centerline{(Bohr magneton for nucleus)} | \centerline{(Bohr magneton for nucleus)} | ||

− | + | Finally, we plug these in for our expression for the energy: | |

− | + | \begin{align} | |

− | &\sim Z^ | + | \Delta E&\sim\frac{e\hbar}{2m_ec}\frac{Ze}{ a_0^2}Z\alpha\\ |

+ | &\sim Z^4\alpha\cdot {\rm Ryd}\\ | ||

+ | \end{align} | ||

+ | 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 |

## Revision as of 13:33, 26 September 2017

### Short Topical Videos

- Radiation and Matter, to Order of Magnitude (Aaron Parsons)
- Bohr Hydrogen Model 1: Radius (Quantum Chemistry)
- Bohr Hydrogen Model 2: Energy (Quantum Chemistry)
- Rotational Spectroscopy Example (Quantum Chemistry)

### Reference Material

## 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:

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 Maxwell’s equation:

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 .