Question

A ruby in a laser consists mostly of alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) and a small amount of chromium ions, responsible for its red color. A \(3.00-\mathrm{kW}\) ruby laser emits light pulses of duration \(10.0 \mathrm{~ns}\) and wavelength \(694 \mathrm{nm}\). a) What is the energy of each of the photons in a pulse? b) Determine the number of chromium atoms undergoing stimulated emission to produce a pulse.

Short answer

Answer: Approximately \(1.04 \times 10^{14}\) chromium atoms undergo stimulated emission to produce a pulse in the ruby laser.

Step-by-step solution

(a) Calculate the energy of each of the photons in a pulse

To find the energy of a photon, we can use Planck's equation: \(E=hf\), where \(E\) is the energy, \(h\) is the Planck's constant, and \(f\) is the frequency of the light. But we are given the wavelength instead, so let's use the speed of light (\(c\)) to find the frequency: \(f=\frac{c}{\lambda}\), where \(\lambda\) is the wavelength. First, let's convert the wavelength to meters: \(694 \textrm{ nm} = 694 \times 10^{-9} \textrm{ m}\). Now, we substitute the values of Planck's constant \((h = 6.63 \times 10^{-34} \textrm{ Js})\) and the speed of light \((c = 3\times10^8 \textrm{ m/s})\) into the equation. \(f=\frac{3\times10^8 \textrm{ m/s}}{694 \times 10^{-9} \textrm{ m}} = 4.33\times10^{14} \textrm{ Hz}\) Now we can compute the photon energy: \(E=hf=(6.63 \times 10^{-34} \textrm{ Js})( 4.33\times10^{14} \textrm{ Hz})= 2.87\times10^{-19} \textrm{ J}\) So, the energy of each photon in a pulse is \(2.87\times10^{-19} \textrm{ J}\).

(b) Determine the number of chromium atoms undergoing stimulated emission to produce a pulse

First, let's compute the pulse energy: \(E_\textrm{pulse} = P \times t\), where \(E_\textrm{pulse}\) is the total energy of the pulse, \(P\) is the power of the laser, and \(t\) is the duration of a pulse. \(E_\textrm{pulse}=(3.00\times10^3 \textrm{ J/s})(10.0\times10^{-9} \textrm{ s})=3\times10^{-5} \textrm{ J}\) Now, we'll determine the number of chromium atoms undergoing stimulated emission to produce a pulse. We can divide the total energy of the pulse by the energy of a single photon to find the number of atoms. \(n =\frac{E_\textrm{pulse}}{E_\textrm{photon}}=\frac{3\times10^{-5} \textrm{ J}}{2.87\times10^{-19} \textrm{ J}}=1.04 \times 10^{14}\) So, there are approximately \(1.04 \times 10^{14}\) chromium atoms undergoing stimulated emission to produce a pulse.