Photon emitted by a hydrogen atom in Lyman series and Balmer series at once ?

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SUMMARY

The discussion centers on the emission of photons by a hydrogen atom during transitions in the Balmer and Lyman series. When a hydrogen atom transitions from the n=3 or n=4 energy levels to n=2, it emits a photon in the Balmer series, followed by a transition to n=1, resulting in a photon emission in the Lyman series. The wavelength of the photon emitted during the Lyman transition is calculated to be 122 nm, derived from the Lyman series equation: 1/λ = R(1 - 1/n²).

PREREQUISITES
  • Understanding of atomic energy levels in hydrogen
  • Familiarity with the Rydberg formula for spectral lines
  • Knowledge of the electromagnetic spectrum, specifically the visible and ultraviolet regions
  • Basic principles of photon emission and absorption in quantum mechanics
NEXT STEPS
  • Study the Rydberg formula in detail for various elements
  • Learn about quantum transitions and their implications in spectroscopy
  • Explore the differences between the Balmer and Lyman series in greater depth
  • Investigate applications of spectral lines in astrophysics and chemical analysis
USEFUL FOR

Students of quantum mechanics, physics educators, and anyone interested in atomic spectroscopy and the behavior of hydrogen atoms during electronic transitions.

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Homework Statement


Whenever a photon is emitted by hydrogen atom in the Balmer series, it is followed by a photon in Lyman series . What wavelength does this latter photon correspond to ?

Homework Equations



Balmer series corresponds to the wavelengths in the visible spectrum and Lyman corresponds to ones in the Ultraviolet.
the equations for each are
Lyman series : 1/λ = R(1 - 1/n2) n = 2,3,4...
Balmer series : 1/λ = R(1/4 - 1/n2) n = 3,4,5...

The Attempt at a Solution



Alright I know that an atom will emit light when it transcends from one state to the other and the energy of the photon is equal to the energy difference between the two states, so if a hydrogen atom emits photon it will emit one which is equal to the energy difference between two state since the energy of that photon is unique so it will correspond to a single wavelength , how can there be another ? This is how it attempted the solution and got stuck at this question .

The answer to this question is : 122nm
 
Physics news on Phys.org
When the H atom has a Balmer transition, the electron ends up at the n=2 energy level.
It then makes an additional transition to n=1.
Applying the Lyman equation to the transition from n=2 to n=1 gives 122nm
 

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