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)

  $\newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}$ $\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}$ $\newcommand {\btau}{\mbox{\boldmath$\tau$}}$ $\newcommand {\bmu}{\mbox{\boldmath$\mu$}}$ $\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}$ $\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}$ $\newcommand {\bomega}{\mbox{\boldmath$\omega$}}$ $\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}$