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SPECTRA PowerPoint Presentation

SPECTRA

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SPECTRA

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  1. SPECTRA

  2. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 N.B. All energies are NEGATIVE. REASON: The maximum energy is the energy to ionise the electron. However an ionised electron feels no attraction to the nucleus so it must have zero potential energy. It follows that energies less than the ionisation energy must be negative e-

  3. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 If a hydrogen atom has its electron in the lowest energy level (ie -13.6eV) it is said to be in the ground state. If an electron absorbs energy and moves to a higher energy level (ie an excited state) it will be unstable and quickly fall back to the ground state, releasing energy as a photon as it falls. Sometimes the energy of these photons correspond to the energy of visible light. e-

  4. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 n=3 -> n=2 n=3 -> n=1 n=2 -> n=1 n=4 -> n=1

  5. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 (ie 3.4-0.85 = 2.55eV (ie 4.08x10-19J)) n=3 -> n=2 n=3 -> n=1 n=2 -> n=1 n=4 -> n=1

  6. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 (ie 3.4-0.85 = 2.55eV (ie 4.08x10-19J)) n=3 -> n=2 (ie 3.4-1.51= 1.89eV (ie 3.024x10-19J)) n=3 -> n=1 n=2 -> n=1 n=4 -> n=1

  7. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 (ie 3.4-0.85 = 2.55eV (ie 4.08x10-19J)) n=3 -> n=2 (ie 3.4-1.51= 1.89eV (ie 3.024x10-19J)) n=3 -> n=1 (ie 13.6-1.51= 12.09eV (ie 1.93x10-18J)) n=2 -> n=1 n=4 -> n=1

  8. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 (ie 3.4-0.85 = 2.55eV (ie 4.08x10-19J)) n=3 -> n=2 (ie 3.4-1.51= 1.89eV (ie 3.024x10-19J)) n=3 -> n=1 (ie 13.6-1.51= 12.09eV (ie 1.93x10-18J)) n=2 -> n=1 (ie 13.6-3.4= 10.2eV (ie 1.63x10-18J)) n=4 -> n=1

  9. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- Determine the amount of energy that would be released when an electron falls from energy level: n=4 -> n=2 (ie 3.4-0.85 = 2.55eV (ie 4.08x10-19J)) n=3 -> n=2 (ie 3.4-1.51= 1.89eV (ie 3.024x10-19J)) n=3 -> n=1 (ie 13.6-1.51= 12.09eV (ie 1.93x10-18J)) n=2 -> n=1 (ie 13.6-3.4= 10.2eV (ie 1.63x10-18J)) n=4 -> n=1 (ie 13.6-0.85= 12.75eV (ie 2.04x10-18J))

  10. n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 Spectrum λ En eV ionisation e- e- e- e- - 13.6 n = 1 ground state BALMER series VISIBLE

  11. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- The energy difference between levels n=4 & n=2 is 2.55eV ie 4.08x10-19J, and between energy levels n=3 & n=2 is 1.89eV ie 3.024x10-19J. These energy drops correspond to the red and green lines in the hydrogen spectrum. The frequency of these lines can be determined by E = hf

  12. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- 4.08x10-19J 3.024x10-19J Use E = hf, and the values of the energy differences between levels to determine the frequencies of the light emitted.

  13. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- 4.08x10-19J 3.024x10-19J E = hf f =E/h f =4.08x10-19/6.63x10-34 f = 6.15x1014Hz E = 4.08x10-19J h = 6.63 x 10-34Js f = ?

  14. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- 4.08x10-19J 3.024x10-19J E = hf f =E/h f =3.024x10-19/6.63x10-34 f = 4.56x1014Hz E = 3.024x10-19J h = 6.63 x 10-34Js f = ?

  15. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- 4.08x10-19J 3.024x10-19J Calculate the wavelength of these radiations. c = f x  E = 3.024x10-19J h = 6.63 x 10-34Js F3-2 = 4.56x1014Hz E = 4.08x10-19J h = 6.63 x 10-34Js F4-2 = 6.15x1014Hz

  16. En eV ionisation n = ∞ 0 n = 4 - 0.85 - 1.51 n = 3 - 3.4 n = 2 n = 1 ground state - 13.6 e- e- 4.08x10-19J 3.024x10-19J • Calculate the wavelength of these radiations. • c = f x  • 3x108=4.56x1014x • = 6.57x10-7m and • 3x108=6.15x1014x • = 4.88x10-7m E = 3.024x10-19J h = 6.63 x 10-34Js F3-2 = 4.56x1014Hz E = 4.08x10-19J h = 6.63 x 10-34Js F4-2 = 6.15x1014Hz