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Waves & Bohr’s Theory Chapter 7 §1-4. Waves Wavelength, λ, in meters (m) The length of a wave from crest to crest or trough to trough. Frequency, υ, in.

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Presentation on theme: "Waves & Bohr’s Theory Chapter 7 §1-4. Waves Wavelength, λ, in meters (m) The length of a wave from crest to crest or trough to trough. Frequency, υ, in."— Presentation transcript:

1 Waves & Bohr’s Theory Chapter 7 §1-4

2 Waves Wavelength, λ, in meters (m) The length of a wave from crest to crest or trough to trough. Frequency, υ, in inverse seconds (s -1 ) The number of waves that pass in a given amount of time.

3 Light’s Wave Characteristics

4 Speed of light equation Speed of light equation c = speed of light, 3.00x10 8 m/s λ = wavelength, in m ν = frequency, in Hz or s –1

5 Light’s Wave Characteristics What is the frequency of red light having a wavelength of 681 nm? 681 nm = 6.81x10 –7 m 3.0x10 8 m/s = (6.81x10 –7 m)( υ ) υ = 4.41x10 14 s –1

6 Max Planck Quantized Quantized

7 Planck’s Equation n = integer other than zero, no unit, represents the energy level of the atom h = Planck’s constant, 6.63x10 –34 Js ν = frequency, in Hz or s –1 E = energy of a quantum – amount of energy to move an e – from its present energy to its next higher one

8 The Photoelectric Effect Albert Einstein incorporated Planck’s ideas into the explanation of the photoelectric effect. Although the photoelectric effect was first discovered by Heinrich Hertz in 1887, Albert Einstein incorporated Planck’s ideas into the explanation of the photoelectric effect.

9 The Photoelectric Effect Einstein stated that electrons could move within their atoms if a minimum amount of energy were reached.

10 Let’s Practice The blue-green line of the hydrogen atom spectrum has a wavelength of 486 nm. What is the energy of a photon of this light? 486 nm = 4.86x10 –7 m 3.0x10 8 m/s = (4.86x10 –7 m) ( υ ) υ = 6.17x10 14 s –1 E = (1)(6.63x10 –34 Js)(6.17x10 14 s –1 ) E = 4.09x10 –19 J

11 Bohr’s Line Spectra Bohr noticed that elements emitted a line spectrum.

12 Bohr’s Line Spectra He hypothesized that each line in the spectra were created when an electron fell from a higher energy level to a lower one within the atom.

13 Bohr’s Line Spectra Here, the electron of the hydrogen atom is shown moving between the various energy levels.

14 Bohr’s Postulates Bohr felt the need to explain two main issues: Bohr felt the need to explain two main issues: 1 st – If electrons are negative and protons are positive….. 1 st – If electrons are negative and protons are positive…..

15 Bohr’s Postulates Bohr felt the need to explain two main issues: Bohr felt the need to explain two main issues: 2 nd – How are the line spectra being created? 2 nd – How are the line spectra being created?

16 Bohr’s Postulates Postulate #1: Postulate #1: Electrons only have specific energy values and energy levels. Electrons only have specific energy values and energy levels. = the energy of a particular e – energy level = the energy of a particular e – energy level = the Rydberg constant = 2.18x10 –18 J = the Rydberg constant = 2.18x10 –18 J = integral value representing the energy level (principal quantum number = integral value representing the energy level (principal quantum number

17 Bohr’s Postulates Postulate #2: Postulate #2: Electrons become excited by collisions of atoms or absorption of energy – think of anything colored… it absorbs light to later emit (reflect) other colors Electrons become excited by collisions of atoms or absorption of energy – think of anything colored… it absorbs light to later emit (reflect) other colors Electrons can change energy only by going from one energy level to another – making a transition. Electrons can change energy only by going from one energy level to another – making a transition.

18 Bohr’s Postulates Postulate #2: Postulate #2: The electron absorbing energy The electron absorbing energy

19 Bohr’s Postulates Postulate #2: Postulate #2: When an electron falls from a higher energy level to a lower energy level, it emits a photon of light. When an electron falls from a higher energy level to a lower energy level, it emits a photon of light.

20 Bohr’s Postulates Postulate #2: Postulate #2: This energy of this photon can be found by: This energy of this photon can be found by:

21 Bohr’s Postulates Postulate #2: Postulate #2: Remember that the energy of a photon can be determined by: Remember that the energy of a photon can be determined by:

22 Bohr’s Postulates Postulate #2: Postulate #2: Which can be related to wavelength: Which can be related to wavelength:

23 Bohr’s Postulates Postulate #2: Postulate #2: Which leads to the Balmer equation: Which leads to the Balmer equation: This equation can calculate the wavelength of any electron falling to the 2 nd energy level – emitting visible light. This equation can calculate the wavelength of any electron falling to the 2 nd energy level – emitting visible light.

24 Bohr’s Postulates Postulate #2: Postulate #2: By the end of Experiment #13, you should be able to identify the Balmer Series, along with the Paschen Series and the Lyman Series. By the end of Experiment #13, you should be able to identify the Balmer Series, along with the Paschen Series and the Lyman Series.

25 Let’s Practice What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from energy level n = 6 to level n = 3? What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from energy level n = 6 to level n = 3?

26 Let’s Practice What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from energy level n = 6 to level n = 3? What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes a transition from energy level n = 6 to level n = 3? ∆E = E 3 – E 6 = υ = 2.74x10 14 s –1 c = λυ → 3.0x10 8 m/s = λ (2.74x10 14 s –1 ) λ = 1.09x10 –6 m = 1090 nm (Infrared)


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