The red light emitted by a ruby laser has a wavelength of \(694.3 \mathrm{nm} .\) What is the difference in energy between the initial state and final state corresponding to the emission of the light?

Understanding wavelength conversion is essential in laser physics, especially when we are dealing with the characteristics of light emitted from lasers. A laser's wavelength, which is typically measured in nanometers (nm), describes the distance between two consecutive peaks of a wave. It's a key feature that determines the type of light we are observing - be it visible, ultraviolet, or infrared.

In the exercise, we're presented with ruby laser light at a wavelength of 694.3 nm. To use this value in equations and calculations that typically require meters (m), we have to convert it. The conversion is straightforward: since 1 nm is equal to \( 10^{-9} \) meters, we multiply the number of nanometers by \( 10^{-9} \). This makes our wavelength \( 694.3 \times 10^{-9} \, \text{m} \), which is now in a form suitable for further calculations.

The speed of light equation is a fundamental concept in physics, relating the speed of light (\(c\)) to the wavelength (\(\lambda\)) and frequency (\(u\)) of a wave. The speed of light is a constant (\(c = 3.0 \times 10^{8} \, \text{m/s}\)) in a vacuum and one of the most critical constants in physics.

When solving for the frequency of light, we rearrange the speed of light equation to \(u = \frac{c}{\lambda}\), which allows us to find the number of wave cycles that pass a given point per second. In the case of the ruby laser light, using its converted wavelength, we calculated the frequency to be \(4.32 \times 10^{14} \, \text{Hz}\). This frequency is an essential property of the light that will help us determine the energy of the photons emitted by the laser.

Planck's equation is vital for understanding the quantum nature of light. This equation links the energy (\(E\)) of a photon to its frequency (\(u\)) via Planck's constant (\(h = 6.63 \times 10^{-34} \, \text{Js}\)). Plank's equation is written as \(E = h \cdot u\).

It tells us that energy is quantized and that photons of light carry discrete energy packets depending on their frequency. In our exercise, by multiplying the frequency calculated in the previous step by Planck's constant, we obtained the energy of a single photon. The result was \(E = 2.87 \times 10^{-19} \, \text{J}\). This value represents the energy difference between the excited state and the ground state in the ruby that creates the red laser light after an electron drops to a lower energy level, emitting a photon of that particular energy. The concept highlights the particle-like properties of light, an essential aspect of quantum mechanics and laser operation.