Calculate the wavelength (in nanometers) of a photon emitted by a hydrogen atom when its electron drops from the \(n=5\) state to the \(n=3\) state.

When considering the behavior of electrons within an atom, transitions between different energy levels are paramount to understanding atomic physics. Imagine an electron orbiting a hydrogen atom nucleus; it can occupy various energy levels or orbits. These energy levels are quantized, meaning electrons can only exist at specific energy states, not in-between.

An electron in a higher energy state (like the n=5 level mentioned in the exercise) may drop to a lower energy state (in this case, the n=3 level). When this happens, the energy difference between these two states is released in the form of a photon. The amount of energy, \( \Delta E \), is crucial because it directly determines the characteristics of the emitted photon, such as its wavelength.

Quantifying Energy Transitions

Calculating this energy difference involves an understanding of the specific energy associated with each level, which, for a hydrogen atom, can be determined by the formula \( E_n = -13.6 eV \times (1/n^2) \), where \(n\) is the principal quantum number corresponding to different energy levels. By calculating the energy of the initial and final states and taking their difference, we can determine the energy released during the transition.

The Rydberg formula is a powerful equation in the world of spectroscopy, particularly for hydrogen-like atoms. This formula allows us to predict the wavelength of light resulting from electron transitions between energy levels of a hydrogen atom.

The formula itself can be expressed as \( 1/\lambda = R_H \times (1/nf^2 - 1/ni^2) \), where \( R_H \) represents the Rydberg constant for hydrogen (\( 1.097 \times 10^7 m^{-1} \)), \( nf \) is the principal quantum number of the final state, and \( ni \) is the principal quantum number of the initial state. In essence, this equation establishes a link between the energy transition of the electron and the wavelength of the emitted photon.

Conversion to Wavelength

Once we compute the right-hand side of the Rydberg equation, the inverse provides us with the wavelength of the photon in question. This impressively simple relationship underpins much of our understanding of atomic emissions and forms the foundation of spectroscopic techniques.

The wavelength and energy of a photon are intimately connected through a relationship known as the Planck-Einstein relation. A core tenet of quantum mechanics and photonics is that each photon's energy is inversely proportional to its wavelength.

Expressed mathematically, the relationship is given by \( E = \frac{hc}{\lambda} \), where \( E \) is the energy of the photon, \( h \) is the Planck constant, \( c \) is the speed of light in a vacuum, and \( \lambda \) is the photon's wavelength. Essentially, as the wavelength of light increases (such as moving from blue to red), the energy decreases and vice versa.

Significance in Atomic Transitions

This fundamental relationship is pivotal when examining atomic transitions. An electron dropping to a lower energy level emits a photon with a specific wavelength directly corresponding to the energy difference between those energy levels. Understanding this concept is key to not only physics but also to numerous applications like lasers, energy-efficient lighting, and spectroscopy.

The Planck-Einstein relation is a cornerstone concept that reconciles the properties of photons with the quantized nature of electron transitions. Through this equation, we can relate the abstract notion of energy levels in an atom to the more tangible measurement of the wavelength of light.

By equating the energy difference calculated using atomic physics (\( \Delta E \)) with the energy of a photon (\( E \) from \( E=hc/\lambda \) ), we tap into a profound understanding of the quantum world. In the educational exercise, after finding the energy difference in electron volts, we convert it into Joules using the conversion factor (1 eV = \( 1.60219 \times 10^{-19} J \) ) to apply this relation.

From Energy to Wavelength

The Planck-Einstein relation is the conduit through which we convert the energy difference to the photon's wavelength. By rearranging the equation to solve for \( \lambda \) (\( \lambda = \frac{hc}{E} \) ), and inserting the known constants for \( h \) and \( c \) along with our energy difference, we arrive at the photon's wavelength in meters. Finally, to assist readability and use in practical contexts, it is often helpful to convert this wavelength into nanometers (nm), which is a commonly used unit in spectroscopy.