12.11: 2P-1S Transitions in Hydrogen
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let us calculate the rate of spontaneous emission between the first excited state (i.e., n=2) and the ground-state (i.e., n′=1) of a hydrogen atom. Now, the ground-state is characterized by l′=m′=0. Hence, in order to satisfy the selection rules ([e13.133]) and ([e13.134]), the excited state must have the quantum numbers l=1 and m=0,±1. Thus, we are dealing with a spontaneous transition from a 2P to a 1S state. Note, incidentally, that a spontaneous transition from a 2S to a 1S state is forbidden by our selection rules.
According to Section [s10.4], the wavefunction of a hydrogen atom takes the form ψn,l,m(r,θ,ϕ)=Rn,l(r)Yl,m(θ,ϕ),
Now, the energy of the eigenstate of the hydrogen atom characterized by the quantum numbers n, l, m is E=E0/n2, where the ground-state energy E0 is specified in Equation ([e9.56]). Hence, the energy of the photon emitted during a 2P→1S transition is ℏω=E0/4−E0=−34E0=10.2eV.
Finally, according to Equation ([e3.115]), the 2P→1S transition rate is written w2P→1S=ω3d23πϵ0ℏc3,
Δλλ∼ΔE2Pℏω∼4×10−8
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)