Respuesta :
Answer:
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
