CHEM HELP PLEASE?
2020-09-23 11:17 am
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If the photon emitted has a wavelength of 434 nm, what is the value of ni?
回答 (2)
2020-09-23 11:38 am
wavelength, λ = 434 nm = 434 × 10⁻⁹ m
Rydberg constant, R = 1.097 × 10⁷ m⁻¹
n₂ = nᵢ to n₁ = 2 (As light is emitted, nᵢ > 2)
Rydberg equation: 1/λ = R [(1/n₁²) - (1/n₂²)]
1/(434 × 10⁻⁹) = (1.097 × 10⁷) × [(1/2²) - (1/nᵢ²)]
(1/4) - (1/nᵢ²) = 1/[(434 × 10⁻⁹) × (1.097 × 10⁷)]
0.25 - (1/nᵢ²) = 0.21
1/nᵢ² = 0.04
nᵢ = √(1/0.04)
nᵢ = 5
Rydberg constant, R = 1.097 × 10⁷ m⁻¹
n₂ = nᵢ to n₁ = 2 (As light is emitted, nᵢ > 2)
Rydberg equation: 1/λ = R [(1/n₁²) - (1/n₂²)]
1/(434 × 10⁻⁹) = (1.097 × 10⁷) × [(1/2²) - (1/nᵢ²)]
(1/4) - (1/nᵢ²) = 1/[(434 × 10⁻⁹) × (1.097 × 10⁷)]
0.25 - (1/nᵢ²) = 0.21
1/nᵢ² = 0.04
nᵢ = √(1/0.04)
nᵢ = 5
2020-09-23 11:26 am
The Rydberg equation gives the wavelength λ for the transitions:
where
R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and
where
R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and
收錄日期: 2021-04-24 08:02:46
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