Use the Rydberg equation to find the wavelength (in \(\mathrm{nm}\) ) of the photon emitted when an electron in an \(\mathrm{H}\) atom undergoes a transition from \(n=5\) to \(n=2\)

In quantum mechanics, quantum numbers are values that describe the energy levels and characteristics of electrons in an atom. They are essential for understanding electron transitions, like in the Rydberg equation. The principal quantum number, denoted as \(n\), indicates the orbit or energy level of an electron, with \(n=1\) being the closest to the nucleus and higher numbers representing farther levels. For instance, in our problem, the electron moves from \(n=5\) to \(n=2\). The change in these numbers is crucial for calculating the energy and, consequently, the wavelength of the photon emitted.

Wavelength calculation becomes straightforward using the Rydberg equation. This formula expresses the inverse of the wavelength (\(1/\lambda\)) as a function of the Rydberg constant (\(R_H\)) and the principal quantum numbers of the initial (\(n_2\)) and final states (\(n_1\)). In our problem, these values are \(n_2 = 5\) and \(n_1 = 2\). By plugging these into the equation \[ \frac{1}{\lambda} = R_H \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] and solving, students can find the wavelength of the light emitted when an electron transitions between these energy levels. It's important to break down each term inside the parentheses before calculating to avoid mistakes.

Photon emission occurs when an electron moves from a higher energy level to a lower one, releasing energy in the form of light. This process is described by the difference in energy levels of the electron. When an electron in a hydrogen atom transitions from \(n=5\) to \(n=2\), it emits a photon. The energy of this photon is directly related to the wavelength of the emitted light, which can be calculated, as shown in the problem. The wavelength determines the color of the light emitted. For the transition from \(n=5\) to \(n=2\), the calculated wavelength is approximately 434 nm, which falls in the visible spectrum, appearing as a blue-violet light.