ATOMIC MODELS. Models are formulated to fit the available data. Atom was known to have certain size. Atom was known to be neutral.

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1 ATOMIC MODELS Models are formulated to fit the available data Atom was known to have certain size. Atom was known to be neutral. Atom was known to give off electrons. THOMPSON MODEL To satisfy the above conditions Thompson proposed Atom made of positive material with enough electrons embedded to make it neutral. 1

2 RUTHERFORD Made measurements of alpha particles scattered from gold: Found 1. Most alpha particles passed through gold without being scattered.. Some alpha particles were scattered through large scattering angles. Data could not be explained by Thompson Model of Atom. 1. How could alpha particles pass through positive mass?. The force between charges was known to be F q 1 r q

3 With the known charges and this force how could alpha particles be scattered through large angles? RUTHERFORD MODEL Small nucleus all positive mass contained in small center part. Electrons in space surrounding nucleus. SCALE MODEL Nucleus Radius = 1 mm Edge of Atom 10 m 3

4 LINE SPECTRA FIRST CONSIDER WHITE LIGHT THROUGH DIFFRACTION GRATING WHITE LIGHT GRATING RAINBOW COLORS 4

5 BUT WITH HYDROGEN GAS AS SHOWN: FIG 3.6 PAGE 93, T & R 5

6 6

7 657 FIG 3.7 PAGE 93, T & R 7

8 THE ONLY WAVELENGTHS PRESENT IN VISIBLE ARE 657 nm (red) 486 nm (blue-green) 434 nm (violet) 410 nm (violet) 397 nm Closely spaced lines out to 365 nm (series limit) 8

9 BALMER Balmer found that he could predict the wavelength of these lines with one equation: R n where R x10 7 m 1 and n 3,4,5,... etc. 9

10 The Balmer lines are in the visible There are other lines: One set in the ultraviolet Three sets in the infrared All wavelengths can be determined by one equation: R n 1 n RYDBERG EQUATION 10

11 WHERE n 1 = 1 and n =,3,4, gives the uv series n 1 = and n = 3,4,5, gives the visible series n 1 = 3 and n = 4,5,6, gives the 1st IR series n 1 = 4 and n = 5,6,7, gives the nd IR series n 1 = 5 and n = 6,7,8, gives the 3 rd IR series 11

12 The series have names for the physicists who did the early work in that part of the spectrum For n 1 = 1 the series is named for Lyman For n 1 = the series is named for Balmer For n 1 = 3 the series is named for Paschen For n 1 = 4 the series is named for Brackett For n 1 = 5 the series is named for Pfund 1

13 THE BOHR MODEL OF THE ATOM Bohr proposed a modification of the Rutherford model that fit the data all data before and including the Rutherford scattering data. The Bohr Atom Postulates: 1. Only orbits allowed are where angular momentum is integral multiple of h. No atom radiates energy as long as electron is in one of the orbital states. 13

14 DERIVATION OF RYDBERG EQUATION (thus validation of Bohr Model) Angular momemtum = mvr m = mass of electron v = velocity of electron in orbit r = radius of orbit From 1 st postulate mvr h n n = 1,, 3, Solving for r r h n mv (1) 14

15 Electron in orbit accelerates toward center a v r The force causing the acceleration is F e r and 15

16 F ma So e r v m r Solving for v 16

17 v e 4 0 mr () Put this in Eq. 1 above Solve for r r n n h me 0 Subscript n to remind one r for each n! 17

18 For n = 1 h 0 r1 10 me x m a0 a 0 is the Bohr orbit for the hydrogen atom in the lowest energy state. Then the other orbits r n n a0 18

19 Bohr model is nucleus with electrons in only specific allowed orbits! 19

20 ENERGY OF THE ATOM Total energy is KE + PE KE 1 mv PE e 4 r 0 and we found v 0

21 1 mr e v 0 4 So r e mr e m E TotalEnergy r e 0 8 Or n n r e E 0 8

22 and r n n h me 0 Putting this in Total Energy equation E n me h n where n= 1,, 3, For: n= 1 E 1 = x J = ev n= E = x J = ev n=3 E 3 = -.41 x J = ev n=4 E 4 = x J = ev

23 Plot these values on a vertical scale to get Figure 4.16 on page 145 of Thornton and Rex 3

24 4

25 DERIVATION OF RYDBERG EQUATION Consider any two energy levels and E i with n i E j with n j If the atom is in n i state with total energy E i and goes to the to the state n j with total energy E j the atom will have less total energy. The difference in energy will be E E i E j The energy is carried off by a photon with energy 5

26 hf So hf E i E j Or since f c 1 f c E i E j 6

27 and E i me h ni E j 4 me h 1 n j So 4 1 me ch n j n i If 7

28 8 me R 1.097x10 m 3 0ch (and it does) Then for n j = 1 We have the Lyman Equation n j = We have the Balmer Equation n j = 3 We have the Paschen Equation n j = 4 We have the Brackett Equation n j = 5 We have the Pfund Equation STUDY FIGURE 4.16 ON PAGE 145 IN THORNTON AND REX!! 8

Optical Spectroscopy and Atomic Structure. PHYS 0219 Optical Spectroscopy and Atomic Structure 1

Optical Spectroscopy and Atomic Structure PHYS 0219 Optical Spectroscopy and Atomic Structure 1 Optical Spectroscopy and Atomic Structure This experiment has four parts: 1. Spectroscope Setup - Your lab