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

1. Nov 12, 2012

### nishantve1

1. The problem statement, all variables and given/known data
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 ?

2. Relevant 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...

3. 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

2. Nov 12, 2012

### ap123

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