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UrbanXrisis
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What is the energy of a photon when the Hydrogen atom transitions from the n=4 to n=3 energy state?
The energy of a photon emitted during the hydrogen n=4 to n=3 transition is 0.137 eV (electron volts). This is equivalent to 2.08 x 10^-19 joules.
The energy of a photon during the hydrogen n=4 to n=3 transition is calculated using the Rydberg formula, which is E = (13.6 eV) * (1/n1^2 - 1/n2^2), where n1 is the initial energy level (n=4) and n2 is the final energy level (n=3).
The hydrogen n=4 to n=3 transition is significant because it is one of the most commonly observed transitions in hydrogen atoms and is responsible for a spectral line known as the Balmer series. This transition also plays a role in the absorption and emission of light in stars, allowing us to study their composition and temperature.
The energy of a photon decreases during the hydrogen n=4 to n=3 transition, as the electron moves from a higher energy level (n=4) to a lower energy level (n=3). This change in energy is emitted as a photon of light.
Yes, the hydrogen n=4 to n=3 transition can occur in reverse, known as the absorption process. This happens when a photon with the same energy as the difference between the two energy levels is absorbed by a hydrogen atom, causing the electron to move from the lower energy level (n=3) to the higher energy level (n=4).