Question

You measure the emission spectrum of hydrogen gas, and see many discrete wavelengths in the emission, of which, one is 102.6nm. To what initial and final states can this wavelength be attributed?. (hint, use a little trial and error to get the intergers)

Answer 1

Some energy Levels and Transition Lines are shown in figure given below..

Corresponding energy equation of photon emmision is given by..

Ephoton = E0(	1	−	1	)
	n12		n22

Where,

E₀ = -13.6 eV

Given,   = 102.6 nm

So, Energy corresponding to emmision of photon with wavelength 102.6 nm is E_photon.

where,

  

By substituting value of we got

E_photon = 12.09 eV

1.) Let electron coming from state 2 to 1

From energy level diagram shown above we got,

E₁ - E₂= - 10.2 eV    E_photon

Negative because electron is coming in lower energy state from higher one by releasing photon of equivalent energy.

So, energy released by H-atom in this case is 10.2 which isn't equal to  E_photon.

It's isn't state 2 to 1 jump.

2.) Let electron coming from state 3 to 1

E1 - E3= - 12.10 eV   - 12.09 =  E_photon

Again, negative because electron is coming in lower energy state from higher one by releasing photon of equivalent energy

It's nearly equal to energy corresponding to a emmited photon with wavelength of 102.6 nm.

Hence for the given data and conditions, electron is undergoing jump from Energy state 3 to Energy state 1 .