Lecture 11: The Bohr Model. PHYS 2130: Modern Physics Prof. Ethan Neil

Transcription

1 Lecture 11: The Bohr Model PHYS 2130: Modern Physics Prof. Ethan Neil

2 Announcements/reminders Midterm Exam #1 next Thursday at 7:30 PM HALE 270 (see class website for details) Midterm covers: Chapter 37 (especially Rutherford + atomic spectra) Chapter 38 (photoelectric matter waves Bohr model) A MIDTERM Reading: review for exam! Other material from lecture (EM waves interference) We will have a review in class next Wednesday. Send me your questions! 2 PHYS 2130 Fall 2017

3 o E free e mmmm. Last time we were studying Bohr s atomic model. Ez E Crucial modification to Rutherford model: electrons exist only in certain discrete energy levels. E f ground state M Discrete energy levels > only discrete energy jumps are allowed! For transitions involving photons we can look for the spectrum. my ; n +0 it it # TO to 3 absorption emission collision PHYS 2130 Fall 2017

4 CQ: A certain type of atom has the energy levels shown below. The atom starts in the ground state. We put the atom in a discharge lamp where an 8 ev electron collides with it. How many different light colors can the atom produce after the collision? A. No light is produced B. 1 color C. 2 colors D. 3 colors ; n +0 to # +0 : Fled 2 ev= E max ij E. 4 colors collision (not an active CQ review) 4 Lecture 10: Atomic Structure PHYS 2130 Fall 2017

5 (not an active CQ review) CQ: A certain type of atom has the energy levels shown below. The atom starts in the ground state. We illuminate the atom with white light and measure the colors of light which are not absorbed. How many different light colors are missing from the spectrum we measure? A. No light is produced B. 1 color C. 2 colors D. 3 colors my +0 it TO : Fled ij E. 4 colors absorption 5 Lecture 10: Atomic Structure PHYS 2130 Fall 2017

6 (example: sodium) E free e o mmmm. Ez E absorption E emission f ground state Important distinction: emission for any n > n but absorption for only 1 > n (excited atoms go very rapidly to the ground state) 6 PHYS 2130 Fall 2017

7 E u ^ binding energies. > r "!. E or state If i= f / ^ ground e E Ez ionization energy z Binding energy is the energy required to remove an electron in a given state from the atom completely. The binding energy of the ground state is called the ionization energy. 7 PHYS 2130 Fall 2017

8 So Bohr s model fixes up the Rutherford atom nicely: Atoms are stable Discrete absorption and! : ; emission spectra explained Every element can have different set of energy levels > unique spectrum But where do these rules come from? Why are electron energies quantized? (How can we predict the allowed energy levels of a given atom?) 8 PHYS 2130 Fall 2017

9 Bohr s theory of the hydrogen atom 9 PHYS 2130 Fall 2017

10 (Beware: some possible naming confusion here ) Bohr s atomic model sometimes refers to the general set of rules we just went over But sometimes it refers to the explicit model of the hydrogen atom we re about to study. I m going to call the latter Bohr s theory of the hydrogen atom to distinguish it from the more general Bohr model. Knight just calls it the Bohr hydrogen atom. Bohr s model is just a set of rules: his theory of hydrogen attempts to explain them at least in part. 10 PHYS 2130 Fall 2017

11 Answering the questions: why are the electron energy levels quantized? Can we predict them? Start with hydrogen the simplest possible atom: one proton + one electron. (If we can t predict that we have no hope for other atoms!) Focus on the circular motion of the electron. Key assumption: electron has a de Broglie wavelength! (Particle in a box returns ) 11 PHYS 2130 Fall 2017

12 CQ: An electron moves in a circular orbit with radius r around a proton. The electron has a de Broglie wavelength λ=h/p. The circular orbit gives a boundary condition for the wave. What relation between r and λ does this boundary condition imply? A. r = n (n=123 ) 12 B. r =2 n C. r = n D. 2 r = n E. 4 r = n (vs. angle: wave(0) = wave(2π) ) PHYS 2130 Fall 2017

13 Electron is in a circular orbit which means as a wave it has a boundary condition: the wave has to repeat if we go around one complete orbit. One orbit = integer # of wavelengths: 2 r = n (n = ) or using de Broglie formula 2 r = h m e v n which we can think of as a constraint on the speed of the electron: v = nh 2 m e r 13 PHYS 2130 Fall 2017

14 italics Circular motion: Coulomb force = centripetal force i ke 2 r 2 = m ev 2 r I. \ i i ) v 2 = ke2 m e r + This tells us that e kinetic energy is determined by r: KE = 1 2 m ev 2 = ke2 2r E = KE + PE = ke2 2r ke 2 r = ke2 2r PHYS 2130 Fall 2017

15 Circular motion: v 2 = ke2 m e r Closed wave orbits: v = nh 2 m e r ) n2 h m 2 er 2 = ke2 m e r italics Multiply through by mer 2 : n 2 h m e = ke 2 r ) r n = n 2 h ke 2 m e Only certain orbital radii are allowed!! 15 PHYS 2130 Fall 2017

16 Messylooking quantity in the equation for r is a distance scale called the Bohr radius: a B h ke 2 m e so the orbits exist at fixed distances: r n = a B n 2 Energy in a given orbit E = KE+PE E n = ke2 2r n = ke 2 2a b n 2 Plugging in for the constants: E n = 13.6 ev n 2 (13.6 ev is known as the Rydberg energy.) 16 PHYS 2130 Fall 2017

17 E n = 13.6 ev n 2 n =3 n=2 n : E g.#iee In =3 fee 01 Ez E n =/ E 1 = E 2 = E 3 = 13.6 ev 13.6 ev 2 2 = 3.4 ev 13.6 ev ev 1 1 E = 13.6 ev = 13.6 ev = 10.2 ev 4 1 E 0 1 = 13.6 ev = 13.6 ev = 12.1 ev 9 17 PHYS 2130 Fall 2017

18 We can write down a general formula for the transition wavelength again by plugging in numbers. Wavelength of light emitted in transition from n to n: 1 1 E n 0!n = 13.6 ev n 02 n 2 = hf = hc = ( ev s)( m/s) 13.6 ev 1 n 02 1 n 2 1 = nm 1 n 02 1 n PHYS 2130 Fall 2017

19 Predict a whole spectrum of energy levels. Historically identified as series with n fixed. = nm 1 n 02 1 n 2 Hydrogen energy levels 1 Balmer series (n=2) was first visible spectrum. Transitions to ground state are UV so harder to see. Lyman series (n=1) important for space telescopes hydrogen gas clouds will absorb 1 >2 photons. 19 PHYS 2130 Fall 2017

20 Let s try this out in detail. Q: What are the first three wavelengths in the Balmer series (n=2)? 3!2 = nm = nm !2 = nm nm nm 1 1 5!2 = nm nm red bluegreen blueviolet 20 PHYS 2130 Fall 2017

21 434 nm 486 nm 656 nm Theory matches experiment nicely! (And we predicted the wavelengths from fundamental constants: h me e k.) CQ: Does the Balmer series end within the visible spectrum? 1!2 = nm nm 4 A. Yes (visible light > 390 nm.) B. No it ends in the UV C. No the series never ends = nm 1 n 02 1 n PHYS 2130 Fall 2017