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

AI Thread Summary
A hydrogen atom emits photons during transitions in both the Balmer and Lyman series, with the Balmer series corresponding to visible wavelengths and the Lyman series to ultraviolet wavelengths. When a photon is emitted in the Balmer series, the atom transitions to the n=2 energy level, followed by a transition to n=1, which emits a photon in the Lyman series. The wavelength of this Lyman series photon is calculated using the Lyman equation, resulting in a wavelength of 122 nm. This process highlights that multiple transitions can occur sequentially, leading to the emission of photons at different wavelengths. Understanding these transitions is crucial for studying atomic emissions in hydrogen.
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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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