12.11: 2P-1S Transitions in Hydrogen

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 \(m=0,\,\pm 1\). 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 \[\psi_{n,l,m}(r,\theta,\phi) = R_{n,l}(r)\,Y_{l,m}(\theta,\phi),\] where the radial functions \(R_{n,l}\) are given in Section [s10.4], and the spherical harmonics \(Y_{l,m}\) are given in Section [sharm]. Some straightforward, but tedious, integration reveals that \[\begin{aligned} \langle 1,0,0|x|2,1,\pm 1\rangle &=\pm \frac{2^{\,7}}{3^{\,5}}\,a_0,\\[0.5ex] \langle 1,0,0|y|2,1,\pm 1\rangle &= {\rm i}\,\frac{2^{\,7}}{3^{\,5}}\,a_0,\\[0.5ex] \langle 1,0,0|z|2,1,0\rangle &=\sqrt{2}\, \frac{2^{\,7}}{3^{\,5}}\,a_0,\end{aligned}\] where \(a_0\) is the Bohr radius specified in Equation ([e9.57]). All of the other possible \(2P\rightarrow 1S\) matrix elements are zero because of the selection rules. It follows from Equation ([e13.128]) that the modulus squared of the dipole moment for the \(2P\rightarrow 1S\) transition takes the same value \[\label{e13.139} d^{\,2} = \frac{2^{\,15}}{3^{\,10}}\,(e\,a_0)^{\,2}\] for \(m=0\), \(1\), or \(-1\). Clearly, the transition rate is independent of the quantum number \(m\). It turns out that this is a general result.

Now, the energy of the eigenstate of the hydrogen atom characterized by the quantum numbers \(n\), \(l\), \(m\) is \(E = E_0/n^{\,2}\), where the ground-state energy \(E_0\) is specified in Equation ([e9.56]). Hence, the energy of the photon emitted during a \(2P\rightarrow 1S\) transition is \[\label{e13.140} \hbar\,\omega = E_0/4 - E_0 = -\frac{3}{4}\,E_0 = 10.2\,{\rm eV}.\] This corresponds to a wavelength of \(1.215\times 10^{-7}\) m.

Finally, according to Equation ([e3.115]), the \(2P\rightarrow 1S\) transition rate is written \[w_{2P\rightarrow 1S} = \frac{\omega^{\,3}\,d^{\,2}}{3\pi\,\epsilon_0\,\hbar\,c^{\,3}},\] which reduces to \[w_{2P\rightarrow 1S} = \left(\frac{2}{3}\right)^8\,\alpha^{\,5}\,\frac{m_e\,c^{\,2}}{\hbar} = 6.27\times 10^8\,{\rm s}^{-1}\] with the aid of Equations ([e13.139]) and ([e13.140]). Here, \(\alpha=1/137\) is the fine-structure constant. Hence, the mean life-time of a hydrogen \(2P\) state is \[\tau_{2P} = (w_{2P\rightarrow 1S})^{-1} = 1.6\,{\rm ns}.\] Incidentally, because the \(2P\) state only has a finite life-time, it follows from the energy-time uncertainty relation that the energy of this state is uncertain by an amount \[{\mit\Delta} E_{2P} \sim \frac{\hbar}{\tau_{2P}}\sim 4\times 10^{-7}\,{\rm eV}.\] This uncertainty gives rise to a finite width of the spectral line associated with the \(2P\rightarrow 1S\) transition. This natural line-width is of order

\begin{equation}\frac{\Delta \lambda}{\lambda} \sim \frac{\Delta E_{2 P}}{\hbar \omega} \sim 4 \times 10^{-8}\end{equation}

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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