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Published byNancy Burker Modified over 4 years ago

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e-e- E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation 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

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e-e- E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation 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.

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E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation Determine the amount of energy that would be released when an electron falls from energy level: (a)n=4 -> n=2 (b)n=3 -> n=2 (c)n=3 -> n=1 (d)n=2 -> n=1 (e)n=4 -> n=1

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

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

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

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

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

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

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E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation The energy difference between levels n=4 & n=2 is 2.55eV ie 4.08x10 -19 J, and between energy levels n=3 & n=2 is 1.89eV ie 3.024x10 -19 J. 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 e-e- e-e-

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

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E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation E = 4.08x10 -19 J h = 6.63 x 10 -34 Js f = ? e-e- e-e- 3.024x10 -19 J 4.08x10 -19 J E = hf f =E/h f =4.08x10 -19 /6.63x10 -34 f = 6.15x10 14 Hz

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

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E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation E = 3.024x10 -19 J h = 6.63 x 10 -34 Js F 3-2 = 4.56x10 14 Hz e-e- e-e- 3.024x10 -19 J 4.08x10 -19 J Calculate the wavelength of these radiations. c = f x E = 4.08x10 -19 J h = 6.63 x 10 -34 Js F 4-2 = 6.15x10 14 Hz

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E n eV - 13.6 n = 1 ground state - 1.51n = 3 0 n = ∞ - 3.4 n = 2 - 0.85 n = 4 ionisation E = 3.024x10 -19 J h = 6.63 x 10 -34 Js F 3-2 = 4.56x10 14 Hz e-e- e-e- 3.024x10 -19 J 4.08x10 -19 J Calculate the wavelength of these radiations. c = f x 3x10 8 =4.56x10 14 x = 6.57x10 -7 m and 3x10 8 =6.15x10 14 x = 4.88x10 -7 m E = 4.08x10 -19 J h = 6.63 x 10 -34 Js F 4-2 = 6.15x10 14 Hz

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