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Emisija fotona

Emisija fotona. n. m. E f = E n - E m. 13,6 eV. , m  n. Balmerova formula. R = 1,097 · 10 7 m -1. , m  n. m = 1, n = 2,3,4...- Lymanov niz. m = 2, n = 3,4,5...- Balmerov niz. m = 3, n = 4,5,6...- Paschenov niz. m = 4, n = 5,6,7...- Brackettov niz.

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Emisija fotona

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  1. Emisija fotona n m Ef = En - Em 13,6 eV , m  n

  2. Balmerova formula R = 1,097 ·107 m-1 , m  n m = 1, n = 2,3,4...- Lymanov niz m = 2, n = 3,4,5...- Balmerov niz m = 3, n = 4,5,6...- Paschenov niz m = 4, n = 5,6,7...- Brackettov niz m = 5, n = 6,7,8...- Pfundov niz

  3. Balmerov niz Paschenov niz Brackettov niz Pfundov niz Lymanov niz n  n = 6 n = 5 n = 4 n = 3 n = 1 n = 2

  4. Primjer: Koje valne duljine može zračiti vodikov atom kada se elektron nađe u trećoj stazi? Rješenje: n = 3 , m = 1 1. n = 3  = 1,03·10-7m ,  = 6,58·10-7m , m = 2 a) n = 3 2. ,  = 1,22·10-7m b) n = 2 , m = 1

  5. Zadatak 1: Izračunajte najmanju i najveću valnu duljinu linija (mi M) u Lymanovu nizu. U kojem se dijelu elektromagnetskog spektra nalazi taj niz? Rješenje: M = 1,2210-7 m = 122nm m = 9,110-8 m = 91nm

  6. Zadatak 2: Elektron se u vodikovu atomu giba po stazi s rednim brojem 7. Prelazeći u nižu stazu, atom emitira valnu duljinu 397 nm. Kojem nizu pripada ta valna duljina? Rješenje: n = 7  = 397 nm = 39710-9 m m = ? m = 2 Valna duljina pripada Balmerov nizu.

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