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When a hydrogen atom undergoes an energy level transition, it emits a photon whose energy is equal to the difference between the two energy levels involved in the transition. For a hydrogen atom moving from a higher energy state, denoted as \( n=3 \), to a lower energy state, \( n=1 \), the process starts by finding the energy levels using the formula: \[ E_n = -13.6 \, \text{eV} \times \frac{1}{n^2}. \]Here, \( E_{n=1} \) is \(-13.6 \, \text{eV} \), and \( E_{n=3} \) works out to be \(-1.51 \, \text{eV} \). The energy difference \( \Delta E \) is then calculated:\[ \Delta E = E_{n=1} - E_{n=3} = 12.09 \, \text{eV}. \]This \( \Delta E \) is the energy of the emitted photon. To use this in further calculations for momentum and wavelength, we convert the energy to joules: \[ \Delta E = 12.09 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.937 \times 10^{-18} \, \text{J}. \]

The momentum of a photon is connected to its energy through the formula:\[ p = \frac{E}{c}, \]where \( p \) is the momentum, \( E \) is the energy of the photon in joules, and \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \). Using the previously calculated energy, the momentum \( p \) can be computed:\[ p = \frac{1.937 \times 10^{-18} \, \text{J}}{3 \times 10^8 \, \text{m/s}} = 6.46 \times 10^{-27} \, \text{kg} \cdot \text{m/s}, \]making it evident that the small value reflects the tiny momentum carried by a photon, despite it being a key player in energy transfer during atomic transitions.

The emitted photon's wavelength is an important characteristic that correlates with its energy and momentum. Using the equation relating energy and wavelength, \[ E = \frac{hc}{\lambda}, \]you can solve for the wavelength \( \lambda \) as follows:\[ \lambda = \frac{hc}{E}, \]where:

• \( h = 6.626 \times 10^{-34} \, \text{Js} \) is Planck's constant.
• \( c = 3 \times 10^8 \, \text{m/s}\) is the speed of light.

By plugging in these constants and the energy value,\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{1.937 \times 10^{-18} \, \text{J}} \approx 1.03 \times 10^{-7} \, \text{m}. \]Converted to more manageable units, this becomes \( \lambda \approx 103 \, \text{nm} \). This wavelength falls within the ultraviolet range of the electromagnetic spectrum.

Energy level transitions in hydrogen atoms illustrate the quantized nature of atomic energy levels. Electrons within an atom reside in discrete energy states, denoted by quantum numbers like \( n=1, 2, 3, \) etc. When an electron transitions from a higher energy level \( n=3 \) to a lower energy level \( n=1 \), the energy difference between these levels is released as a photon. The transition involves:

• Absorption of energy: When an electron jumps to a higher energy level.
• Emission of energy: When an electron falls to a lower energy level, releasing a photon.

Such transitions provide valuable insights into the atomic structure and form the basis for many spectroscopic techniques used in modern science.

Planck's constant \( h \) is one of the fundamental constants in quantum physics, essential for characterizing the quantization of energy levels. It acts as a bridge between macroscopic and quantum phenomena.In calculations involving photon energy and wavelength, Planck's constant is employed as:\[ E = \frac{hc}{\lambda} \] and is key to determining how energy and frequency (or wavelength) are related. With its small value of \( 6.626 \times 10^{-34} \, \text{Js} \), Planck's constant reveals that photons, while carrying discrete energy packets, have significant implications on a micro-scale. Its introduction was pivotal in the development of quantum mechanics, reshaping our understanding of atomic and subatomic processes.This concept not only frames spectral line evaluations but also helps define the rules governing energy transactions in atomic transitions, making \( h \) a cornerstone of quantum theory.