7
$\begingroup$

Was wondering what the average time is for an electron on any given orbital, or how often they change energy levels.

Thanks in advance.

$\endgroup$
1
  • $\begingroup$ The part of physics dealing with the motion of electrons in atoms is atomic physics, not nuclear physics. $\endgroup$ Jun 23, 2016 at 7:52

1 Answer 1

8
$\begingroup$

Short answer: of the order of a nanosecond for hydrogen for "allowed" transitions, and the emission rate scales roughly as $Z^2$, where $Z$ is the atomic number. For an oxymoronically named "forbidden" transition, these times increase to tens of milliseconds or fractions of a second.

So let's elaborate: what sets these times?

A point not made enough is that "orbital" is usually taken to mean an energy eigenstate of the Hamiltonian $H_0$ for the electron in a "bare" atom. So, if the "bare" atom were a true model of the atom's nature, an electron in a non-ground-state orbital could not decay at all (the state is an eigenstate of the Hamiltonian), so its lifetime would be infinite!

The decay happens because atom systems are not "bare" like this: these systems are always coupled to quantized electromagnetic field and so the eigenstates of $H_0$ (the orbitals) are not eigenstates of the whole system.

So, to do the full calculation, one needs to know the full Hamiltonian $H_0 + H^\prime$ of combined, coupled atom and EM field. A feeling for this kind of calculation is given in my answer here; the co-efficients in that analysis must in turn emerge from full quantum electrodynamics. But it turns out that a semiclassical analysis for hydrogen yields the right results. One thinks of the excited hydrogen as a classical dipole antenna with dipole moment amplitude $\mu$, which radiates power $P=2\,\omega^4\,\mu^2/(3\,c^3)$ where $\hbar\,\omega$ is the energy level difference for the transition, whence the transition rate in photons per second is $\Gamma = 2\,\omega^3\,\mu^2/(3\,\hbar\,c^3)$. The reciprocal $(3\,\hbar\,c^3)/(2\,\omega^3\,\mu^2)$ estimates how long the dipole takes to radiate its photon. We need now to estimate the dipole moment. This is done (rather crudely) by multiplying the electron's charge $q$ by the mean distance between the electron in the two orbitals; this expectation is $e\,\langle \psi_1\mid \vec{x}\mid\psi_2\rangle$. Amazingly, plugging this value into our classical rate estimate in fact gives us the exact transition rate as calculated from QED; it is:

$$\Gamma_{1\to2} = \frac{4\,\omega^3\,q^2}{3\,\hbar\,c^3} \left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2$$

For example, for the $2\,p\to1\,s$ transition, this formula gives a rate of $6.25\times 10^8{\rm s^{-1}}$, or a lifetime of about $1.4{\rm ns}$. For the $3\,s\to2\,p$ with a much, much smaller overlap $\left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2$ the formula gives $6.3\times 10^6{\rm s^{-1}}$, or a very long lifetime of about $150{\rm ns}$.

The above are all "allowed" transitions, where the overlap $\left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2$ is nonzero. However, all orbitals have a definite parity, being odd or even under spatial inversion; this means that $\psi(-\vec{x}) = \pm\psi(\vec{x})$, so that the quantity $\left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2$ can only be nonzero if one of the orbitals has even, the other odd symmetry to cancel out the odd symmetry of the $\vec{x}$ operator in middle.

"Forbidden" transitions between orbitals of the same parity still happen, but they are much, much slower: they must decay in a two stage process, each stage flipping the parity, where there are two photons whose energy sums to the transition energy. For example, state $2\,s$ lands in the opposite parity (but higher energy) intermediate $2\,p$ state flipping the parity on a "borrowed" quantity of energy and then "pays back" the energy in making the second parity flipping transition $2\,p\to1\,s$. The transition time for this process is $0.12{\rm s}$.

$\endgroup$

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.