Chapter 29 Atomic Physics. Looking Ahead. Slide 29-1

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1 Chapter 29 Atomic Physics Looking Ahead Slide 29-1

2 Atomic Spectra and the Bohr Model In the mid 1800s it became apparent that the spectra of atomic gases is comprised of individual emission lines. Slide 29-2

Atomic Spectra and the Bohr Model In 1885 the four visible lines of atomic hydrogen were accurately measured by Anders Angstrom Johann Balmer found that the four

3 Atomic Spectra and the Bohr Model In 1885 the four visible lines of atomic hydrogen were accurately measured by Anders Angstrom Johann Balmer found that the four lines could be fit by the formula = R 4 n 2 R = nm -1 (Rydberg Constant) Balmer predicted other lines exist in the UV for higher values of n. Slide 29-3

4 Rydberg Formula Johannes Rydberg generalized the formula to = R n 02 n 2 (n >n 0 ) The Balmer series are for the case n =2 If n =3, photons are in the infrared. In 1908 Louis Paschen observed some of them. If n =1, photons are in the UV. In 1914 Theodore Lyman observed some of these. Example: n =1, n=2 (Lyman alpha transition) 1 = R(1 1/4) = 3 4 R = nm Slide 29-4

5 Thomson s Model of the Atom Thomson s Plumb Pudding model: individual point-like electrons are embedded in a uniform distribution of positive charge. Slide 29-5

6 Rutherford Scattering 1909 Rutherford & Geiger devised a scattering experiment to probe the structure of atoms. They used a radioactive source of alpha particles (helium nuclei) and a thin metal foil. Geiger & Masder investigated how many alpha particles scatter as a function of angle, including large angles Slide 29-6

Force on alpha particle due to positive charge Electrons are much less massive than an alpha particle, thus as an alpha particle plows through the atom, the electrons are ejected (without deflecting

7 Force on alpha particle due to positive charge Electrons are much less massive than an alpha particle, thus as an alpha particle plows through the atom, the electrons are ejected (without deflecting the alpha particle) Thus we only have to worry about the positive charge Electric field due to positive charge depends on radius. Field is strongest near surface Alpha particles that skim surface should scatter the most (about 1 degree) rutherford-scattering PHET Slide 29-7

8 Rutherford s Model 1910 Geiger and Marsden found that some alpha particles found with a scattering angle over 140 degrees! HIGHLY improbable based on Thomson s atomic model This result led Rutherford to come up with his nuclear model of the atom (1911). The only way for an alpha particle to backscatter is to have a very strong electric force Thus the positive charge must be very small. In fact, to fit the data, the nucleus had to be less than m! The atom is 99.99% empty space! Most mass in nucleus. Slide 29-8

scattering as a function of scattering angle is N( ) = Nnd Zke 2 2 1 4s

9 Confirmation of Rutherford s Model Using Newton s second law and the Coulomb force, Rutherford predicted that that the number of particles scattering as a function of scattering angle is N( ) = Nnd Zke s 2 E sin 4 ( /2) Perfect agreement with Geiger and Marsden s 1913 data Slide 29-9

10 Problems With Rutherford s Model Rutherford s model of hydrogen was an electron orbiting around a nucleus like a planet orbiting a star. This model predicts atoms are very unstable Accelerating electron emits radiation, loses energy, and spirals inward (in a timescale of s!!). This model predicts a continuous spectrum This model can not explain why the spectrum contains discrete lines Unclear how nuclei with more than one proton stays Slide 29-10

11 Bohr s Model of the Hydrogen Atom Bohr s model (1913) was an attempt to understand the emission lines and the stability of the hydrogen atom Bohr proposed the existence of stationary states (or stationary orbits). The electron can orbit the proton only in one of these discrete states. Bohr couldn t explain why the allowed states are discrete or why electrons in these states do not radiate. These stationary states have different energies. The atom can emit (or absorb) photons such that the change in energy between the states is equal to the energy of the emitted (or absorbed) photon. E n E n 0 = = hc Slide 29-11

12 Bohr s Model For an electron in orbit around the proton, the electron s centripetal acceleration is due to the Coulomb force: F = ke2 r 2 = m e This implies that the electron s kinetic energy is 1 2 m ev 2 = 1 2 ke 2 v 2 The total energy of the electron is then Bohr s quantization condition: l = mvr = n h 2 = n~ n~ ) m e v 2 = m e m e r r r 2 = ke2 r E = 1 2 ke 2 r ) r = n2 ~ 2 ke 2 m e Slide 29-12

13 This gives the allowed orbital radii: r = n2 ~ 2 ke 2 m e = a B n 2 The Bohr radius is a B = ~2 ke 2 m e = nm Now that we got r, we can get the energies of the stationary states: E n = 1 ke 2 = 1 ke 2 2 r n 2 a B n 2 E n = Or, E R n 2, where E R = m(ke2 ) 2 2~ 2 = 13.6 ev Slide 29-13

14 Bohr s Model of Atomic Quantization An atom can undergo a transition or quantum jump from one stationary state to another by emitting or absorbing a photon whose energy is exactly equal to the energy difference between the two stationary states Pearson Education, Inc. Slide 29-14

15 Atomic Transitions Consider what happens when a hydrogen atom makes a transition from a stationary state with a high energy (say state n) to a lower energy state (state n ). The difference in the energy levels is 1 1 E = E n 0 E n = E R n 02 n 2 To conserve energy, a photon is emitted with an energy = hc = ) 1 = E R hc E 1 n 02 1 n 2 E R hc = R Perfect agreement with Rydberg s equation! Slide 29-15

16 Question Atomic Transitions II The Balmer series for hydrogen can be 1) 3 2 observed in the visible part of the 2) 4 2 spectrum. Which transition leads to 3) 5 2 the reddest line in the spectrum? 4) 6 2 5) 2 n = n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 Slide 29-16

17 Slide 29-17

18 Understanding Emission and Absorption Lines Sun s spectrum Slide 29-18

19 QuickCheck 29.2 An atom has the energy levels shown. A photon with a wavelength of 620 nm has an energy of 2.0 ev. Do you expect to see a spectral line with wavelength of 620 nm in this atom s absorption spectrum? A. Yes B. No C. There s not enough information to tell Pearson Slide 29-19

20 Question Energy Levels II The emission spectrum for the atoms of a gas is shown. Which of the energy level diagrams below corresponds to this spectrum? (1) (2) (3) (4) Slide 29-20

21 QuickCheck 29.3 An atom has the energy levels shown. A beam of electrons with 5.5 ev kinetic energy collides with a gas of these atoms. How many spectral lines will be seen? A. 2 B. 3 C. 4 D. 5 E Pearson Slide 29-21

22 Successes of Bohr Model 'Explains' source of Balmer formula and predicts empirical constant R (Rydberg constant) from fundamental constants: R= 1 / 91.2 nm =mk 2 e 4 /(4πc! 3 ) Explains why R is different for different single electron atoms (called hydrogen-like ions). Predicts approximate size of hydrogen atom Explains (sort of) why atoms emit discrete spectral lines Explains (sort of) why electron doesn t spiral into nucleus Nobel Prize in 1922 Slide 29-22

23 Shortcomings of the Bohr model: Does not explain why the orbital angular momentum of the electron is quantized. Does not explain why electrons don t radiate when they are in stationary orbitals despite the fact that Coulomb s law still works. Can t explain spectra of multi-electron atoms Can t explain or predict the relative intensities of emission lines (thus can t explain the cause of the relative probabilities between atomic transitions). Can t explain Zeeman effect and line splitting due to internal magnetic fields. Slide 29-23

De Broglie s Explanation of Bohr s Quantization De Broglie envisioned the electron in a hydrogen atom as somehow vibrating around the Bohr orbit, similar to a standing

24 De Broglie s Explanation of Bohr s Quantization De Broglie envisioned the electron in a hydrogen atom as somehow vibrating around the Bohr orbit, similar to a standing wave of a guitar string. In order for the wave to to fit without overlapping, 2 r = n = n h p ) l = rp = n~ This is Bohr s quantization of angular momentum! Slide 29-24

25 Solutions to the Schrodinger eqn for Hydrogen 1. Schrödinger found that the energy of the hydrogen atom is given by the same expression found by Bohr: The integer n is called the principle quantum number. 3. The wavefunction for the electron yield a 3D probability cloud around the nucleus. 2. The angular momentum L of the electron s orbit must be one of the values The integer l is called the orbital quantum number. 3. The component of the angular momentum in the z-direction is given by L z = mħ, where l m l. Slide 29-25

26 Hydrogen Atom Slide 29-26

27 Slide 29-27

28 Slide 29-28

29 The Schrodinger Equation The wave formulation of quantum mechanics is a VERY successful theory. Some of the phenomena it is able to account for are: Energy states of atoms, and relative intensities of emission lines due to atomic transitions Structure of the periodic table Molecular orbitals and all of physical chemistry The phenomena of quantum tunneling Theory of solid-state physics, transistors, diodes, semiconductors, etc. Alpha decay of heavy nuclei and energy states of nuclei Slide 29-29

Chapter 28. Atomic Physics

Chapter 28. Atomic Physics Chapter 28 Atomic Physics Sir Joseph John Thomson J. J. Thomson 1856-1940 Discovered the electron Did extensive work with cathode ray deflections 1906 Nobel Prize for discovery of electron Early Models