NIVEL ENERGIA DO ATOMO DE BOHR

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𝐸 = 𝐸𝑐 + 𝐸𝑝 𝐸=

1 π‘˜π‘’ 2 π‘šπ‘£ 2 + 2 2 π‘Ÿ

Quando o eletron movendo no nucleo a forΓ§a coulomb Γ© igual a forΓ§a centripetal 𝐹𝑒 = 𝐹𝑐 π‘˜π‘’ 2 π‘šπ‘£ 2 = π‘Ÿ2 π‘Ÿ Para obter a formula energia total eletron multiplicamos Β½ ambos os lados 1 π‘˜π‘’ 2 π‘šπ‘£ 2 1 π‘˜π‘’ 2 π‘šπ‘£ 2 π‘˜π‘’ 2 π‘šπ‘£ 2 π‘₯ = π‘₯ β†’ 2= β†’ = 2 π‘Ÿ2 π‘Ÿ 2 2π‘Ÿ 2π‘Ÿ 2π‘Ÿπ‘Ÿ 2π‘Ÿ π‘˜π‘’ 2 𝐸𝑛 = βˆ’ 2π‘Ÿ Onde π‘Ÿ = o raio de Bohr 𝑛 2 ℏ2 π‘Ÿπ‘› = β†’ π‘Ÿπ‘› = 𝑛2 π‘Ÿπ΅ 4Ξ 2 π‘šπ‘˜π‘’ 2 𝐸𝑛 = βˆ’

π‘˜π‘’ 2 2π‘Ÿ

π‘˜π‘’ 2 𝐸𝑛 = βˆ’ 2 2𝑛 π‘Ÿπ΅ Onde o raio de igual 0,5292 𝐸𝑛 = βˆ’

(9π‘₯109 )π‘₯(1,60π‘₯10βˆ’19 )2 2π‘₯0,5292 𝑛2

𝐸𝑛 = βˆ’

2,304π‘₯10βˆ’28 0,1058 𝑛2

𝐸𝑛 = βˆ’

2,17769π‘₯10βˆ’27 𝑛2

1 𝐽 = 1,60π‘₯10βˆ’19 𝑒𝑉 𝐸𝑛 = βˆ’

2,17769π‘₯10βˆ’27 1,60π‘₯10βˆ’19 𝑒𝑉𝑛2

𝐸𝑛 = βˆ’

13,6 𝑒𝑉 𝑛2


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