Was wondering what the average time is for an electron on any given orbital, or how often they change energy levels.
Thanks in advance.
Answer
Short answer: of the order of a nanosecond for hydrogen for "allowed" transitions, and the emission rate scales roughly as Z2, 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 H0 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 H0 (the orbitals) are not eigenstates of the whole system.
So, to do the full calculation, one needs to know the full Hamiltonian H0+H′ 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 μ, which radiates power P=2ω4μ2/(3c3) where ℏω is the energy level difference for the transition, whence the transition rate in photons per second is Γ=2ω3μ2/(3ℏc3). The reciprocal (3ℏc3)/(2ω3μ2) 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⟨ψ1∣→x∣ψ2⟩. Amazingly, plugging this value into our classical rate estimate in fact gives us the exact transition rate as calculated from QED; it is:
Γ1→2=4ω3q23ℏc3|⟨ψ1∣→x∣ψ2⟩|2
For example, for the 2p→1s transition, this formula gives a rate of 6.25×108s−1, or a lifetime of about 1.4ns. For the 3s→2p with a much, much smaller overlap |⟨ψ1∣→x∣ψ2⟩|2 the formula gives 6.3×106s−1, or a very long lifetime of about 150ns.
The above are all "allowed" transitions, where the overlap |⟨ψ1∣→x∣ψ2⟩|2 is nonzero. However, all orbitals have a definite parity, being odd or even under spatial inversion; this means that ψ(−→x)=±ψ(→x), so that the quantity |⟨ψ1∣→x∣ψ2⟩|2 can only be nonzero if one of the orbitals has even, the other odd symmetry to cancel out the odd symmetry of the →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 2s lands in the opposite parity (but higher energy) intermediate 2p state flipping the parity on a "borrowed" quantity of energy and then "pays back" the energy in making the second parity flipping transition 2p→1s. The transition time for this process is 0.12s.
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