The Bohr atom and the Uncertainty Principle Previous Lecture: Matter waves and De Broglie wavelength The Bohr atom This Lecture: More on the Bohr Atom The H atom emission and absorption spectra Uncertainty Principle Wave functions 1
MTE 3 Contents, Apr 23, 2008 Ampere s Law and force between wires with current (32.6, 32.8) Faraday s Law and Induction (ch 33, no inductance 33.8-10) Maxwell equations, EM waves and Polarization (ch 34, no 34.2) Photoelectric effect (38.1-2-3) Matter waves and De Broglie wavelength (38.4) Atom (37.6, 37.8-9, 38.5-7) Wave function and Uncertainty (39) Requests for alternate exams should be sent before Apr. 20 2
HONOR LECTURE PROF. RZCHOWSKI COLOR VISION 3
Swinging a pendulum: the classical picture Larger amplitude, larger energy Small energy Large energy d d Potential Energy E=mgd E.g. (1 kg)(9.8 m/s2 )(0.2 m) ~ 2 Joules
The quantum mechanics scenario Energy quantization: energy can have only certain discrete values Energy states are separated by ΔE = hf. f = frequency h = Planck s constant= 6.626 x 10-34 Js d Suppose the pendulum has Period = 2 sec Frequency f = 0.5 cycles/sec ΔE min =hf=3.3x10-34 J << 2 J Quantization not noticeable at macroscopic scales
Bohr Hydrogen atom Hydrogen-like atom has one electron and Z protons the electron must be in an allowed orbit. Each orbit is labeled by the quantum number n. Orbit radius = n 2 a o /Z and a0 = Bohr radius = 0.53 A The angular momentum of each orbit is quantized Ln = n The energy of electrons in each orbit is En = -13.6 Z 2 ev/n 2 6
Radius and Energy levels of H-like ions Charge in nucleus Ze (Z = number of protons) Coulomb force centripetal F = k Ze2 = m v 2 r 2 r mvr = nh h 2 r = n 2 = n 2 mkze 2 Z a 0 Total energy: E = p2 2m kze2 r E = 1 2 k 2 Z 2 me 4 = 13.6eV Z 2 n 2 h 2 n 2 The minimum energy (binding energy) we need to provide to ionize an H atom in its ground state (n=1) is 13.6 ev (it is lower if the atom is in some excited state and the electron is less bound to the nucleus). Correspondence principle : quantum mechanics must agree with classical results when appropriate (large quantum numbers) r = n 2 a 7 0
F = k Ze2 = m v 2 r 2 r mvr = nh Substitute (2) in (1): r = k Ze2 m 2 r 2 r = n 2 mn 2 h 2 Z r = n 2 Z a 0 Put numbers in h 2 me 2 k = n 2 Z v = nh mr k Ze2 r = mv 2 (1) r = k Ze2 mv 2 Total energy = kinetic energy + potential energy = kinetic energy + ev E = p2 2m k Ze2 r E = Z 2 = mv 2 2 k Ze2 r (2) (1.055 10 34 ) 2 9.11 10 31 (1.6 10 19 ) 2 9 10 9 = n 2 = k Ze2 2r n 2 (9 10 9 ) 2 9.11 10 31 (1.6 10 19 ) 4 2 (1.055 10 34 ) 2 = Z 2 Z=n. of protons in nucleus of H-like atom n 13.6eV 2 8 Z 0.53 10 10 m a 0 = Bohr radius = 0.53 x 10-10 m = 0.53 A Ze2 = k 2 Zme2 k = Z 2 k 2 me 4 = 13.6eV Z 2 n 2 h 2 n 2 2h 2 n 2 n 2 2.2 10 18 J = Z 2
Photon emission question An electron can jump between the energy levels of on H atom. The 3 lower levels are part of the sequence En = -13.6/n 2 ev with n=1,2,3. Which of the following photons could be emitted by the H atom? A. 10.2 ev B. 3.4 ev C 1.0 ev E 3 =-1.5 ev 13.6 3.4 =10.2eV 13.6 1.5 =12.1eV 3.4 1.5 =1.9eV E 2 =-3.4eV ev E 1 =-13.6 ev 9
Energy conservation for Bohr atom Each orbit has a specific energy E n =-13.6/n 2 Photon emitted when electron jumps from high energy Eh to low energy orbit El (atom looses energy). E l E h = h f <0 Photon absorption induces electron jump from low to high energy orbit (atom gains energy). E l E h = h f >0 Bohr successfully explained H-like atom absorption and emission spectra but not heavier atoms 10
Example: the Balmer series All transitions terminate at the n=2 level Each energy level has energy E n =-13.6 / n 2 ev E.g. n=3 to n=2 transition Emitted photon has energy E photon = 13.6 13.6 3 2 2 2 =1.89 ev Emitted wavelength E photon = hf = hc λ, λ = hc E photon = 1240 ev nm 1.89 ev = 656 nm 11
Hydrogen spectra Lyman Series of emission lines: Balmer: 1 λ = R 1 2 1 2 n 2 n=3,4,.. n=2,3,4,.. R = 1.096776 x 10 7 /m Use E=hc/λ Rydberg-Ritz Hydrogen For heavy atoms R = 1.097373 x 10 7 /m 12
Suppose an electron is a wave Here is a wave: λ = h p x λ where is the electron? Wave extends infinitely far in +x and -x direction
Analogy with sound Sound wave also has the same characteristics But we can often locate sound waves E.g. echoes bounce from walls. Can make a sound pulse Example: Hand clap: duration ~ 0.01 seconds Speed of sound = 340 m/s Spatial extent of sound pulse = 3.4 meters. 3.4 meter long hand clap travels past you at 340 m/s
Beat frequency: spatial localization What does a sound particle look like? One example is a beat frequency between two notes Two sound waves of almost same wavelength added. Constructive interference Large amplitude Destructive interference Small amplitude Constructive interference Large amplitude
Creating a wave packet out of many waves Tbeat = Δt = 1/Δf =1/(f2-f1) f1 = 440 Hz f2= 439 Hz Soft Loud Soft Loud f1+f2 Beat 440 Hz + 439 Hz + 438 Hz 440 Hz + 439 Hz + 438 Hz + 437 Hz + 436 Hz Construct a localized particle by adding together waves with slightly different wavelengths (or frequencies).
A non repeating wave...like a particle Six sound waves with slightly different frequencies added together We do not know the exact frequency: Sound pulse is comprised of several frequencies in the range Δf, we are uncertain about the exact frequency The smallest our uncertainty on the frequency the larger Δt, that is how long the wave packet lasts Δf Δt 1 8 4 0-4 -8 Δt -15-10 -5 0 5 10 15 J
8 4 Heisenberg s Uncertainty principle The packet is made of many frequencies and wavelengths The uncertainty on the wavelength is Δλ de Broglie: λ = h /p => each of the components has slightly different momentum. During Δt the wave packet length is 0-4 -8-15 -10-5 0 5 10 15 Δx J The uncertainty in the momentum is: f = v λ = p x m p x h = p 2 x mh Δf = 2p x hm Δp x x Δx = vδt = p x m Δt this the uncertainty on the particle location Δf Δt 1 Δf Δt = 2p x hm Δp x m p x Δx = 2 h Δp x Δp x Δx h 2 Δx 1
Uncertainty principle The more precisely the position is determined, the less precisely the momentum is known in this instant, and vice versa. --Heisenberg, uncertainty paper, 1927 Δp x Δx h 2 Particles are waves and we can only determine a probability that the particle is in a certain region of space. We cannot say exactly where it is! Similarly since E = hf ΔE Δt h 2 19
Uncertainty principle question Suppose an electron is inside a box 1 nm in width. There is some uncertainty in the momentum of the electron. We then squeeze the box to make it 0.5 nm. What happens to the momentum? A. Momentum becomes more uncertain B. Momentum become less uncertain C. Momentum uncertainty unchanged 20
Electron Interference superposition of 2 waves with same frequency and amplitude D = D 1 + D 2 = Asin(kr 1 ωt) + Asin(kr 2 ωt) Intensity on screen and probability of detecting electron are connected Computer simulation x πdx I(x) = Ccos 2 λl I A 2 photograph
Wave function When doing a light interference experiment, the probability that photons fall in one of the strips around x of width dx is P(x)dx = = I(x)Hdx N tot hf N(in dx at x) N tot A(x) 2 dx Energy(in dx at x) /t N tot hf /t = The probability of detecting a photon at a particular point is directly proportional to the square of lightwave amplitude function at that point P(x) is called probability density (measured in m -1 ) P(x) A(x) 2 A(x)=amplitude function of EM wave Similarly for an electron we can describe it with a wave function ψ(x) and P(x) ψ(x) 2 is the probability density of finding the electron at x H 22 x
Wave Function of a free particle Ψ(x) must be defined at all points in space and be single-valued Ψ(x) must be normalized since the particle must be somewhere in the entire space The probability to find the particle between xmin and xmax is: Ψ(x) must be continuous in space There must be no discontinuous jumps in the value of the wave function at any point P(x min x x max ) = x max x min Ψ(x) 2 dx 23
Where is most propably the electron? your HW How do I get a? from normalization condition area of triangle =(a x 4)/2 = 1 Suppose now that a is another number, a = 1/2, what is the probability that the electron is between 0 and 2? A. 1/2 B. 1/4 C. 2 Probability = area of tringle between 0,2 = (a x 2)/2 = 1/2 24
Wave Function of a free particle Ψ(x) may be a complex function or a real function, depending on the system For example for a free particle moving along the x-axis Ψ(x) = Ae ikx k = 2π/λ is the angular wave number of the wave representing the particle A is the constant amplitude Remember: complex number imaginary unit 25