41.8.3 The Helium-Neon Laser

The familiar red laser used in lecture demonstrations, laboratories, and supermarket checkout scanners is the helium-neon laser, often called a HeNe laser. Its output is a continuous, rather than pulsed, wavelength of 632.8 nm. The medium of a HeNe laser is a mixture of $\approx 90 %$ helium and $\approx 10 %$ neon gases. As FIGURE 41.35a shows, the gases are sealed in a glass tube, then an electrical discharge is established along the bore of the tube. Two mirrors are bonded to the ends of the discharge tube, one a total reflector and the other having $\approx 2 %$ transmission so that the laser beam can be extracted.

The figure labeled (“a”) illustrates the elements of a helium neon laser.

The figure labeled (b) illustrates the excitation transfer of helium atoms to neon.

Full Alternative Text

The atoms that lase are the neon atoms, but the pumping method involves the helium atoms. The electrons in the discharge collisionally excite the $1 s 2 s$ state of helium. This state has a very low spontaneous decay rate (i.e., a very long lifetime) because a decay back to the $1 s^{2}$ state would violate the $Δ l$ selection rule, so it is possible to build up a fairly large population (but not an inversion) of excited helium atoms in the $1 s 2 s$ state. The energy of the $1 s 2 s$ state is 20.6 eV.

Interestingly, an excited state of neon, the $5 s$ state, also has an energy of 20.6 eV. If a $1 s 2 s$ excited helium atom collides with a ground-state neon atom, as frequently happens, the excitation energy can be transferred from one atom to the other! Written as a chemical reaction, the process is

{He}^{*} + Ne \to He + {Ne}^{*}

where the asterisk indicates the atom is in an excited state. This process, excitation transfer, is very efficient for the $5 s$ state because the process is resonant—a perfect energy match. Thus the two-step process of collisional excitation of helium, followed by excitation transfer between helium and neon, pumps the neon atoms into the excited $5 s$ state. This is shown in FIGURE 41.35b.

The $5 s$ energy level in neon is $\approx 1.95 eV$ above the $3 p$ state. The $3 p$ state is very nearly empty of population, both because it is not efficiently populated in the discharge and because it undergoes very rapid spontaneous emission to the $3 s$ states. Thus the large number of atoms pumped into the $5 s$ state creates a population inversion with respect to the lower $3 p$ state. These are the necessary conditions for laser action.

Because the lower level of the laser transition is normally empty of population, placing only a small fraction of the neon atoms in the $5 s$ state creates a population inversion. Thus a fairly modest pumping action is sufficient to create the inversion and start the laser. Furthermore, a HeNe laser can maintain a continuous inversion and thus sustain continuous lasing. The electrical discharge continuously creates $5 s$ excited atoms in the upper level, via excitation transfer, and the rapid spontaneous decay of the $3 p$ atoms from the lower level keeps its population low enough to sustain the inversion.

A typical helium-neon laser has a power output of $1 mW = 10^{- 3} J / s$ at 632.8 nm in a 1-mm-diameter laser beam. As you can show in a homework problem, this output corresponds to the emission of $3.2 \times 10^{15}$ photons per second. Other continuous lasers operate by similar principles, but can produce much more power. The argon laser, which is widely used in scientific research, can produce up to 20 W of power at green and blue wavelengths. The carbon dioxide laser produces output power in excess of 1000 W at the infrared wavelength of $10.6 μ m$ . It is used in industrial applications for cutting and welding.

EXAMPLE 41.9 An ultraviolet laser

An ultraviolet laser generates a 10 MW, 5.0-ns-long light pulse at a wavelength of 355 nm. How many photons are in each pulse?

SOLVE The energy of each light pulse is the power multiplied by the duration:

E_{pulse} = P Δ t = (1.0 \times 10^{7} W) (5.0 \times 10^{- 9} s) = 0.050 J

Each photon in the pulse has energy

E_{photon} = h f = \frac{h c}{λ} = 3.50 eV = 5.60 \times 10^{- 19} J

Because $E_{pulse} = N E_{photon}$ , the number of photons is

N = E_{pulse} / E_{photon} = 8.9 \times 10^{16} photons

CHAPTER 41 CHALLENGE EXAMPLE Electron probability in hydrogen

What is the probability that a $1 s$ hydrogen electron is found at a distance from the proton that is less than half the Bohr radius?

MODEL The Schrödinger model of the hydrogen atom represents the electron as a wave function. We can’t say exactly where the electron is, but we can calculate the probability of finding it in a specified region of space.

SOLVE We’re interested in finding the electron not at a certain point in space but within a certain distance from the nucleus. For this we use the radial probability density

P_{r} (r) = 4 π r^{2} | R_{n l} (r) |^{2}

where $R_{n l} (r)$ is the radial wave function, rather than the square of the wave function $ψ (x, y, z)$ . The probability of finding the electron at a distance between $r_{min}$ and $r_{max}$ is

\begin{array}{l} Prob (r_{min} \leq r \leq r_{max}) & = & \int_{r_{min}}^{r_{max}} P_{r} (r) d r \\ = & 4 π \int_{r_{min}}^{r_{max}} r^{2} {| R_{n l} (r) |}^{2} d r \end{array}

The $1 s$ radial wave function was given in Equations 41.7:

R_{1 s} (r) = \frac{1}{\sqrt{π a_{B}^{3}}} e^{- r / a_{B}}

where $a_{B}$ is the Bohr radius. We specify that the electron is less than half the Bohr radius from the proton by setting $r_{min} = 0$ and $r_{max} = \frac{1}{2} a_{B}$ . Thus the probability we seek is

\begin{array}{l} Prob (r \leq \frac{1}{2} a_{B}) & = & 4 π \int_{0}^{a_{B} / 2} r^{2} {| R_{1 s} (r) |}^{2} d r \\ = & \frac{4 π}{π a_{B}^{3}} \int_{0}^{a_{B} / 2} r^{2} e^{- 2 r / a_{B}} d r \end{array}

To evaluate this integral, it will be useful to change variables. Let $u = 2 r / a_{B}$ , so that the exponential can be written more simply as $e^{- u}$ . Turning this around, we have $r = \frac{1}{2} a_{B} u$ and thus

r^{2} d r = {(\frac{1}{2} a_{B} u)}^{2} (\frac{1}{2} a_{B} d u) = \frac{1}{8} a_{B}^{3} u^{2} d u

A change of variables requires a corresponding change of limits: When $r = 0, u = 0$ also; when $r = \frac{1}{2} a_{B}, u = 1$ . With these substitutions, the probability calculation becomes

Prob (r \leq \frac{1}{2} a_{B}) = \frac{1}{2} \int_{0}^{1} u^{2} e^{- u} d u

This looks much nicer! Notice that all the $a_{B}$ have disappeared, so our answer will be a numerical value.

This is not an easy integral, but it is a common one. It can be found in integral tables, such as in Appendix A, or evaluated with mathematical software. The result is

\begin{array}{l} Prob (r \leq \frac{1}{2} a_{B}) & = & \frac{1}{2} {[- (u^{2} + 2 u + 2) e^{- u}]}_{0}^{1} \\ = & \frac{1}{2} (2 - 5 e^{- 1}) = 0.080 \end{array}

The probability that a $1 s$ hydrogen electron is less than half the Bohr radius from the proton is 0.080, or 8.0%.

REVIEW The probability is small, but that is not unexpected. The graph of the radial probability density in Figure 41.7 shows that the probability peaks at $r = a_{B}$ and then decreases rapidly as $r \to 0$ . We can see that the area under that curve from $r = 0$ to $r = \frac{1}{2} a_{B}$ is not large. The electron can be found much closer to the proton than one Bohr radius, but not with a large probability.