Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum. View Available Hint(s)

 λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

        Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

                 ED = 13.606 (1 / n₀² - 1 /)

we are going to apply this relationship to answer slash.

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name        Initial state

Lymman         1

Balmer           2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

              ΔE = 13.606 (1/2² - 1 /∞)

              ΔE = 3.4015 eV

Let's use the Planck relation for the energy

                E = h f = h c /λ

                λ = h c / E

Let's reduce the energy to J

              E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

            λ = 6.63 10⁻³⁴  3 10⁸ / 5.4424 10⁻¹⁹

            λ = 3.654 10⁻⁷ m

            λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

            λ =365.4 nm

this eta radiation in the ultraviolet range

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