= 3. The emission spectrum of Hydrogen has a bright blue-green line at 486 nm. This electron ends in the nf 2 energy level. What is the initial energy of the electron that produced this photon? (Find the energy of the electron in its excited state, n₁). m.)

Inquiry into Physics
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ISBN:9781337515863
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Chapter10: Atomic Physics
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Work through the following problem to determine if the answer is correct or incorrect. Explain why. 

3. The emission spectrum of Hydrogen has a bright blue-green line at 486 nm. This electron ends in the nf =
2 energy level. What is the initial energy of the electron that produced this photon? (Find the energy of
the electron in its excited state, nį).
с
Ephoton = h = (6.626 × 10-34 J. s)
Ephoton = 4.09 × 10-1⁹ J = 2.56 eV
Ephoton = -4Eelectron -(Eƒ – E₁)
-13.6 eV
n²
E₁ = Ephoton + Ef = 2.56 eV +
•
(3 × 108 m)
(486 × 10-⁹ m)
E = 2.56 eV +
(-13,6 eV)
E₁ =
= -0.84 eV
Transcribed Image Text:3. The emission spectrum of Hydrogen has a bright blue-green line at 486 nm. This electron ends in the nf = 2 energy level. What is the initial energy of the electron that produced this photon? (Find the energy of the electron in its excited state, nį). с Ephoton = h = (6.626 × 10-34 J. s) Ephoton = 4.09 × 10-1⁹ J = 2.56 eV Ephoton = -4Eelectron -(Eƒ – E₁) -13.6 eV n² E₁ = Ephoton + Ef = 2.56 eV + • (3 × 108 m) (486 × 10-⁹ m) E = 2.56 eV + (-13,6 eV) E₁ = = -0.84 eV
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