Respuesta :
Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted will be : d < a < c < b
[tex]E = -13.6[/tex]×[tex]Z[/tex]²/[tex]n[/tex]²
where,
[tex]E[/tex] = energy of orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
[tex]E[/tex]₁ = -13.6× 1²/1² = -13.6eV
Energy of n = 2 in an hydrogen atom:
[tex]E[/tex]₂ = -13.6 × 1²/2² = -3.40eV
Energy of n = 3 in an hydrogen atom:
[tex]E[/tex]₃ = -13.6× 1²/3² = -1.51eV
Energy of n = 4 in an hydrogen atom:
[tex]E[/tex]₄ = -13.6× 1²/4² = -0.85eV
Energy of n = 5 in an hydrogen atom:
[tex]E[/tex]₃ = -13.6× 1²/ 5² = -0.54eV
a) n = 2 to n = 4 (absorption)
Δ[tex]E[/tex]₁ = E₄ - E₂ = -0.85- (- 3.40) = 2.55eV
b) n = 2 to n = 1 (emission)
Δ E₂ = E₁ - E₂ = -13.6 - (- 3.40) = -10.2eV
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
ΔE₃ = E₅ - E₂ = -0.54 -( -3.40) = -2.85eV
d) n = 4 to n = 3 (emission)
ΔE₄ = E₃ - E₄ = -1.51 - (- 0.85) = - 0.66eV
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
[tex]E = hv[/tex]
h = Planck's constant
[tex]v =[/tex]frequency of the wave
So, the increasing order of magnitude of the energy difference :
[tex]E[/tex]₄< E₁ <E₃ <E₂
The H atom electron transitions in order of increasing frequency of the photon absorbed or emitted will be d < a < c < b
: d < a < c < b
To learn more about transitions visit the link:
https://brainly.com/question/28304182?referrer=searchResults
The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
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