3
$\begingroup$

I read Griffiths EM today and it says something very interesting but a little bothering to me. It states for an atom, the position of center of mass of an electron cloud lies in the center of the proton.

But my question is why? What will happen if this is not true?

$\endgroup$
0

2 Answers 2

3
+50
$\begingroup$

The electron bound to a proton should not be thought of as a point particle. Its position should be thought of as a superposition of all possible electron positions, weighted by their probability. The object which describes the likelihood of observing the electron at any one position or momentum is the wavefunction $\Psi$, which is a function of the coordinates of space, $r,\theta,\phi$.

The electron bound inside a hydrogen atom has a wavefunction that obeys the following Hamiltonian $$\left[-\frac{1}{2\mu}\frac{\partial^2}{\partial r^2} - \frac{2}{r}\frac{\partial}{\partial r} + \frac{\ell(\ell + 1)}{2\mu r^2} - \frac{e^2}{r}\right]\psi(r) = E \psi(r)\,.$$ Here, we've separated the electron wavefunction into a radial part $\psi(r)$ and an angular part $Y_{\ell m}(\theta,\phi)$, so the total wavefunction is written $\psi(r)Y_{\ell m}(\theta,\phi)$. $\ell$ is a quantum number that represents the amount of orbital angular momentum carried by the electron. The coordinate $r$ represents the separation of the proton and the electron, and $\mu$ is the reduced mass of the electron and proton $$\mu = \frac{m_e m_p}{m_e + m_p}\approx m_e + \mathscr{O}(m_e^2/m_p)\,.$$ The Hamiltonian consists of four terms, although the first two should be thought of as a single term representing the radial kinetic energy of the electron-proton system. The third term represents the fact that two bodies with mutual angular momentum will not be able to collide head-on. The fourth term is the attractive electric potential between the proton and electron.

In nature, the electron-proton system will tend to occupy the lowest energy level accessible to it. In this case, the $\ell = 0$ state, corresponding to a spherically symmetric wavefunction, is the lowest energy configuration of the system. Up to a normalization factor, the ground state is $$\Psi(r,\theta,\phi) \propto e^{-r/2r_b}\,.$$ Here, $r_b$ is the bohr radius of the atom. We are now in a position to address your question.

In order to ask "where is the electron?" we need to be a bit more precise in what we mean. The electron can in principle be anywhere, since the electron wavefunction is non-zero everywhere. Therefore, the correct question to ask is "where am I most likely to find the electron?" The job of answering this question is given to a set of objects called "expectation values," typically written as $\langle x \rangle$ where $x$ is the quantity you are measuring and $\langle\cdot\rangle$ represents an ensemble average. In quantum mechanics, the expectation value is computed as follows $$\langle x\rangle = \frac{\int{\rm d} V \Psi^* x \Psi}{\int{\rm d} V \Psi^*\Psi}$$ In our case, we're interested in measuring the expectation value of the position operator $\vec x$. Therefore, we must compute (up to a normalization) $$\langle \vec x\rangle = \int{\rm d} V \Psi^* \vec x \Psi = \int{\rm d} V \left[\begin{array}{c}r \sin\theta\cos\phi\\r \sin\theta\sin\phi\\ r\cos\theta\end{array}\right] e^{-r/r_b}\,.$$ Performing the angular integrals yields the result $$\langle \vec x\rangle = \vec 0\,.$$ Recalling the physical meaning of the $r$ coordinate is the separation of the proton and electron, we find that the most probable position of the electron is precisely on top of the proton. In other words, the center of mass of the electron is aligned with the center of mass of the proton.

Another question you might ask is "how far away from the proton should the electron be on average?" Now we're interested in the expectation value of the operator $r$, instead of $\vec x$. $$\langle r \rangle =\frac{ \int{\rm d} V \Psi^* r \Psi}{ \int{\rm d} V \Psi^*\Psi} = \frac{\int{\rm d} V r e^{-r/r_b}}{\int{\rm d} V e^{-r/r_b}} = r_b\,.$$ Thus, we expect to find the electron a Bohr radius away from the proton!

To summarize: the electron will most likely be found at the origin, but its typical separation from the proton is a Bohr radius.

$\endgroup$
-1
$\begingroup$

The electron cloud is not exacly centered at the proton in the hydrogen atom (or about the nucleus in other atoms).

This is analogous to the Moon not exactly orbiting about the Earth, but the Earth and the Moon each orbiting about their barycenter (center of mass of the Earth - Moon System).

In solving the Schrodinger equation for the hydrogen atom, it is modeled that the proton is fixed, and the reduced mass of the electron, instead of the actual mass in used. This does not mean the proton is actually fixed relative to the orbitals of the electron. However, since the proton is 1836 times heavier than the electron, it is a good approximation to say the electron cloud is centered about the proton (or nucleus).

For further information see this reference.

$\endgroup$
7
  • 1
    $\begingroup$ It isn't modeled that proton is fixed in solving of Schrödinger equation: the problem can be solved with arbitrary finite nuclear mass by changing variables from $r_p$ and $r_e$ to $r_p-r_e$ and $r_p+r_e$. In this form the usual solution is the center-of-mass solution (wavefunction depends on separation here), and the wavefunction cusp is precisely at proton. $\endgroup$
    – Ruslan
    Mar 29, 2014 at 18:52
  • $\begingroup$ Also, the Moon does orbit the Earth: the barycenter of Earth-Moon system is about 1710 km under Earth's surface. That's pretty deep: Earth radius is only about 6400 km. So, I'd say your answer is wrong in almost all aspects... $\endgroup$
    – Ruslan
    Mar 29, 2014 at 19:01
  • $\begingroup$ @Ruslan The question specifically refers to "the center of the proton", so I am considering that small deviations analogous to the moon orbiting a point ~4700km off of the center of the Earth to be relevant. $\endgroup$
    – DavePhD
    Mar 30, 2014 at 0:28
  • $\begingroup$ @Ruslan "Quantum Chemistry", by McQuarrie, begins the solution to the Schrodinger equation for the hydrogen atom with the statement "As our model, we shall picture the hydrogen atom as a proton fixed at the origin, and an electron of reduced mass...". I am saying that the electron orbitals are actually about the center of mass, which is slightly different than the center of the proton. In other words, that the electron and proton both move about the center of mass of the system. Would you disagree? $\endgroup$
    – DavePhD
    Mar 30, 2014 at 0:39
  • 1
    $\begingroup$ Yes, I disagree. See Landau, Lifshitz "Quantum mechanics. Non-relativistic theory", ch. 32 "Motion in a centrally symmetric field". There L&L don't do any simplifications as McQuarrie does in your textbook, and instead they separate the Hamitonian into sum of Hamiltonians for motion of barycenter and two-body problem. Then the solution is $\psi(r)Y(\theta,\phi)e^{i\vec k\vec R}$, where $r$ is separation, $\theta, \phi$ are angles of barycentric spherical coordinates, and $\vec R$ is radius-vector of barycenter. From this form it's obvious that barycenter is at the point of full collision. $\endgroup$
    – Ruslan
    Mar 30, 2014 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.