Ahydrogen atom is excited from its ground state to then = 4 state. the atom subsequently emits three photons, one of which has a wavelength of 122 nm .

what are the wavelengths of the other two photons?

enter your answers in ascending order separated by commas.

   102 <122 <658 nm

Explanation:

The energy levels of the hydrogen atom are described by the Bohr model

        Eₙ = -13.606 / n²        [eV]

        n = 1, 2, 3, …

Where n is an integer

For an atom excited to the level n = 4

Let's look for the energy of these states

Let's use the Planck equation and the speed of light to find the wavelength

        E = h f

        c = λ f

        E = h c / λ

        λ = h c / E

Reductions

        1eV = 1.6 10⁻¹⁹ J

Let's start

n = 1

       E₁ = -13,606 eV

n = 2

      E₂ = -13.606 / 2² = -3.4015 eV

n = 3

      E₃ = -13.606 / 3² = -1.5118 eV

n = 4

      E₄ = -13.606 / 4² = - 0.8504 eV

Let's look for the wavelength of a specific energy

To pass the energy of eV to Joule multiply by 1.6 10-19 J

       λ = 6.63 10⁻³⁴ 3 10⁸ / E 1.6 10⁻¹⁹

       λ = 12.43125 10⁻⁷/E

Let's reduce the meter wavelength to nm 1 m = 10⁹ nm

The transitions are

State             DE = Ef-Ei (eV) lam (10-7 m) lam (nm)

Last  initial

4          1       -0.8504 - (-13606)  = 12.756     0.975       97.5

4           2      -0.8504 - (-3.4015) = 2.551       4.87       487

4           3      -0.8504 - (- 1.5118) = 0.661       18.80    1880

3           1       -1.5118 - (-13,606)  = 12,094       1.02       102

3           2      -1.5118 - (-3.4015)  = 1.890         6.58      658

2           1      -3.4015- (-13.606)  = 10.205       1.22       122

If the transition is to state n = 3, the transitions are

The energy of the transition in ascending order is

        12,094> 10,205> 1,890

The wavelengths are

          102 <122 <658 nm

no. dot product   =0 when vecs at right angles