The number of B2H6 molecules
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Hence the order of the transition in order of increasing frequency of the photon absorbed or emitted will be : d < a < c < b
×
²/
²
where,
= energy of orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
₁ = -13.6× 1²/1² = -13.6eV
Energy of n = 2 in an hydrogen atom:
₂ = -13.6 × 1²/2² = -3.40eV
Energy of n = 3 in an hydrogen atom:
₃ = -13.6× 1²/3² = -1.51eV
Energy of n = 4 in an hydrogen atom:
₄ = -13.6× 1²/4² = -0.85eV
Energy of n = 5 in an hydrogen atom:
₃ = -13.6× 1²/ 5² = -0.54eV
a) n = 2 to n = 4 (absorption)
Δ₁ = 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:
h = Planck's constant
frequency of the wave
So, the increasing order of magnitude of the energy difference :
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:
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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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