Question 42.11: CASE STUDY Got to Pump In order for a helium–neon laser to o......
CASE STUDY Got to Pump
In order for a helium–neon laser to operate, the neon atoms must be in an excited state. The wavelength of light from a helium–neon laser is 633 nm. At normal room temperature, what is the ratio of the number density of neon atoms in the high-energy state to that of the atoms in the low-energy state?
Interpret and Anticipate
We can use the wavelength of the emitted light to find the energy difference between the two energy states that emitted that light. Then we can use Boltzmann’s distribution law to find the ratio.
Solve
The energy of a photon can be found by combining f = c/λ (Eq. 34.20) and E = hf (Eq. 40.8).
The energy of the photon must equal the difference in the energy of the two states.
\begin{aligned}& E_{ photon }=E_{ hi }-E_{ lo }=\Delta E \\& \Delta E=\frac{h c}{\lambda}=\frac{\left(6.63 \times 10^{-34} J \cdot s \right)\left(3.00 \times 10^8 m / s \right)}{633 \times 10^{-9} m } \\& \Delta E=3.14 \times 10^{-19} J\end{aligned}To use Boltzmann’s law (Eq. 42.31), we must also estimate room temperature. Typically, room temperature is about 20° C = 293 K. Rearrange to find the ratio and substitute values.
\begin{aligned}& n_{ hi }=n_{ lo } e^{-\Delta E / k_{ B } T } \quad \quad (42.31) \\& \frac{n_{ hi }}{n_{ lo }}=e^{-\Delta E / k_{ B } T } \\& \frac{n_{ hi }}{n_{ lo }}=\exp \left(\frac{-3.14 \times 10^{-19} J }{\left(1.38 \times 10^{-23} J \cdot K \right)(293 K )}\right) \\& \frac{n_{ hi }}{n_{ lo }}=1.78 \times 10^{-34}\end{aligned}Check and Think
We just found a very small ratio. To make this more understandable, if there were 1 mole of neon gas, and so there were 6.0 \times 10^{23} atoms, then no atom would likely be in the high-energy state. Now we see why pumping and the existence of a metastable state are so important. Stimulated emission requires atoms in the higher energy state. Atoms must be pumped to the higher state and remain there until they can be simulated to emit photons.