Problem 62 - Fundamentals of Physics

Question

Ruby lases at a wavelength of \(694 \mathrm{~nm}\). A certain ruby crystal has \(4.00 \times 10^{19} \mathrm{Cr}\) ions (which are the atoms that lase). The lasing transition is between the first excited state and the ground state, and the output is a light pulse lasting \(2.00 \mu\) s. As the pulse begins, \(60.0 \%\) of the \(\mathrm{Cr}\) ions are in the first excited state and the rest are in the ground state. What is the average power emitted during the pulse? (Hint: Don't just ignore the ground-state ions.)

Step-by-Step Solution

Answer

The average power emitted is approximately 3.43 MW.

Step 1: Calculate Energy of Each Photon

The energy of each photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ J s} \), \( c \) is the speed of light \( 3.00 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength \( 694 \times 10^{-9} \text{ m} \). Plugging the values into the equation gives us:\[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{694 \times 10^{-9}}\approx 2.86 \times 10^{-19} \text{ J/photon}\]

Step 2: Determine Number of Excited Ions Initially

Determine the number of \( \mathrm{Cr} \) ions initially in the excited state. We know that \( 60.0\% \) of the \( 4.00 \times 10^{19} \) ions are in the excited state:\[\text{Number of excited ions} = 0.60 \times 4.00 \times 10^{19} = 2.40 \times 10^{19} \text{ ions}\]

Step 3: Calculate Energy Released During Pulse

The total energy emitted is the product of the number of excited ions transitioning and the energy per photon. Thus, the total energy released during the laser pulse is:\[E_{\text{total}} = 2.40 \times 10^{19} \times 2.86 \times 10^{-19} \approx 6.86 \text{ J}\]

Step 4: Calculate Average Power

The average power \( P \) is the total energy divided by the duration of the pulse. Since the pulse lasts \( 2.00 \mu \text{s} \), which is \( 2.00 \times 10^{-6} \text{ s} \):\[P = \frac{E_{\text{total}}}{\text{pulse duration}} = \frac{6.86}{2.00 \times 10^{-6}} \approx 3.43 \times 10^6 \text{ W} \]Thus, the average power emitted during the pulse is approximately \( 3.43 \times 10^6 \text{ watts} \).

Key Concepts

Photon Energy Calculation

Understanding how to calculate photon energy is crucial in laser physics. Photon energy is the energy carried by a single photon with a given wavelength. We use the formula:\[ E = \frac{hc}{\lambda} \]where:

• \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \, \text{J s} \).
• \( c \) is the speed of light, \( 3.00 \times 10^8 \, \text{m/s} \).
• \( \lambda \) is the wavelength, \( 694 \times 10^{-9} \, \text{m} \) for a ruby laser.

For a photon emitted by a ruby laser, substituting these values gives:\[ E \approx 2.86 \times 10^{-19} \, \text{J/photon} \]. This small amount of energy is significant because it represents the energy transition between excited energy states of an atom.
In laser technology, multiple such photons combine to yield a powerful laser beam, each contributing a quantum of energy.

Excited State Population

When discussing lasers, one key concept is the population of excited states in atoms or ions. In a ruby laser, chromium ions (Cr) are essential for achieving the necessary energy states. Initially, a percentage of these ions are excited to a higher energy state, ready to release energy as laser light.
In our example, 60% of the chromium ions—among \( 4.00 \times 10^{19} \) ions—are excited, which means:

• Calculated number of excited ions: \[ 0.60 \times 4.00 \times 10^{19} = 2.40 \times 10^{19} \text{ ions} \].

This means that a majority of the ions contribute to the laser emission process. The population inversion, where more ions are in the excited state than the ground state, is essential for achieving laser action.
The greater the population of excited ions, the higher the potential energy released, since they all transition back to the lower energy state, emitting photons in the process.

Laser Pulse Duration

The duration of a laser pulse is another important factor in laser operation. The pulse duration refers to the time over which the laser emits energy. In this context, our pulse duration is \(2.00 \, \mu\text{s}\) or \(2.00 \times 10^{-6} \, \text{s}\).
This short duration is typical for laser pulses, which are brief yet intense. Such short bursts of energy allow lasers to emit light without continuous operation.Understanding how this duration affects power output is crucial:

• The average power \( (P) \) of the laser is calculated by dividing the total emitted energy by the pulse duration using the relation: \[ P = \frac{E_{\text{total}}}{\text{pulse duration}} \].

Given a total energy of approximately \( 6.86 \, \text{J} \) and pulse duration, the power generated is \(3.43 \times 10^6\) watts.
This shows how even short pulses can release considerable power, aided by population inversion and photon energy, facilitating applications ranging from medical procedures to industrial cutting.