Question

A ruby 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. One such laser of power \(3.00 \mathrm{~kW}\) emits light pulse of duration \(10.0 \mathrm{~ns}\) and of wavelength \(685 \mathrm{nm}\). a) What is the energy of the photons in the pulse? b) Determine the number of chromium atoms undergoing stimulated emission to produce this pulse.

Short answer

Answer: The energy of the photons in the pulse is \(2.90 \times 10^{-19}\mathrm{J}\) and the number of chromium atoms undergoing stimulated emission to produce this pulse is approximately \(1.03 \times 10^{14}\) atoms.

Step-by-step solution

Calculate the frequency of the light

To find the energy of the photons, we first need to determine the frequency of the light. We can use the relationship between the speed of light, wavelength, and frequency: \(f = \dfrac{c}{\lambda}\) where c = 3.00 × 10^8 m/s (speed of light) and λ = 685 nm (wavelength) Convert wavelength to meters: \(\lambda = 685 \times 10^{-9} \mathrm{m}\) Now, plug in the values and calculate the frequency: \(f = \dfrac{3.00 \times 10^{8}\mathrm{m/s}}{685 \times 10^{-9}\mathrm{m}}\) \(f = 4.38 \times 10^{14}\mathrm{Hz}\)

Calculate the energy of the photons

Now, we can use the formula \(E = hf\) to calculate the energy of the photons: \(E = h \times f\) where h = 6.626 × 10^-34 Js (Planck's constant), and f = 4.38 × 10^14 Hz Now, plug in the values and calculate the energy of the photons: \(E = (6.626 \times 10^{-34}\mathrm{Js})(4.38 \times 10^{14}\mathrm{Hz})\) \(E = 2.90 \times 10^{-19}\mathrm{J}\)

Calculate the energy of the pulse

Using the given power and duration, we can find the energy of the pulse using the formula \(E_{pulse} = P \times t\): \(E_{pulse} = P \times t\) where P = 3.00 kW (power) and t = 10.0 ns (duration) Convert power and duration to SI units: \(P = 3.00 \times 10^{3}\mathrm{W}\), \(t = 10.0 \times 10^{-9}\mathrm{s}\) Now, plug in the values and calculate the energy of the pulse: \(E_{pulse} = (3.00 \times 10^{3}\mathrm{W})(10.0 \times 10^{-9}\mathrm{s})\) \(E_{pulse} = 3.00 \times 10^{-5}\mathrm{J}\)

Calculate the number of chromium atoms undergoing stimulated emission

Finally, we can find the number of chromium atoms undergoing stimulated emission using the formula \(N = \dfrac{E_{pulse}}{E_{photon}}\): \(N = \dfrac{E_{pulse}}{E_{photon}}\) where \(E_{pulse} = 3.00 \times 10^{-5}\mathrm{J}\) and \(E_{photon} = 2.90 \times 10^{-19}\mathrm{J}\) Now, plug in the values and calculate the number of chromium atoms undergoing stimulated emission: \(N = \dfrac{3.00 \times 10^{-5}\mathrm{J}}{2.90 \times 10^{-19}\mathrm{J}}\) \(N \approx 1.03 \times 10^{14}\) So, the energy of the photons in the pulse is \(2.90 \times 10^{-19}\mathrm{J}\) and the number of chromium atoms undergoing stimulated emission to produce this pulse is approximately \(1.03 \times 10^{14}\) atoms.

Key concepts

Photon Energy

When we talk about photon energy, we are referring to the energy carried by a single photon, the basic unit of light. Photons are tiny, but their energy can be quantified through a well-known equation:

• The energy of a photon is given by the formula: \[ E = hf \]where \( E \) is the energy, \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{Js})\), and \( f \) is the frequency of the light.
• To calculate the frequency \( f \), we can use the speed of light \( c \) and the wavelength \( \lambda \). This relationship is expressed as:\[ f = \dfrac{c}{\lambda} \]with \( c = 3.00 \times 10^{8} \text{m/s} \) and \( \lambda \) converted to meters.

These calculations facilitate our understanding of how much energy a single photon can hold, which is essential in laser physics. This determines how effective the laser can be for different applications, such as cutting materials or in medical procedures.

Stimulated Emission

Stimulated emission is a core concept in laser physics describing how lasers amplify light. It was first proposed by Albert Einstein and is fundamental in explaining how lasers work:

• When an atom or ion in an excited state (higher energy) encounters a photon with the right energy, it can be stimulated to emit another photon of the same energy and phase.
• This emitted photon is coherent with the incident photon — they share the same wavelength and direction, resulting in the amplification of light.
• In a laser, this process is repeated as photons stimulate other excited atoms, leading to the exponential growth of light.

By harnessing stimulated emission, lasers can emit highly focused and intense light beams, useful in various scientific and industrial applications. This process is what makes lasers unique compared to other light sources.

Chromium Ions

Chromium ions, specifically \( \text{Cr}^{3+} \), play a vital role in ruby lasers. Let's understand their function:

• Ruby lasers are made from aluminum oxide, \( \text{Al}_2\text{O}_3 \), doped with a small amount of chromium ions.
• These ions are responsible for the red color of the laser light due to their unique absorption and emission characteristics.
• When these chromium ions absorb energy, they move from a lower energy state to an excited state. Upon returning to the lower energy state, they release photons which contribute to the laser's output.

The understanding of chromium ions' role is essential for optimizing laser performance and customizing lasers for specific tasks, such as precise drilling or cutting, based on their output characteristics.