The Structure of the Atom Chapter 4 in Rex and Thornton s Modern Physics Wed. October 26, 2016 S 1
In this chapter... S We ll explore how our understanding of atomic structure developed S Ancient Greek philosopher Democritus thought that all of matter was composed of an indivisible object called the a-tom S Not to be confused with the modern atom, which is divisible S Investigations into the structure of the atom accelerated (1900s) S With the discovery of many elements, and then the electron (which was presumed to be related of atoms) at the end of the 19 th century 2
Competing Models S The Thomson Model S Electrons are distributed over a volume, much like raisins in a pudding (which is positively charged) S In this model, if you heat an element or apply a potential, you could cause electron(s) to vibrate -> giving rise to EM radiation. S However, this model could not be used to calculate to line spectra for hydrogen S The Rutherford Model S In his experiments, Rutherford shot alpha particles (He atoms with their electrons stripped off, but we did not know that) at a very thin gold foil, and he observed what happened S Where did the α s go 3 aka the plum-pudding model
Contrast S If the Thomson model were correct, then the alpha particle could hit one of the electrons and scatter. S Example 4.1 gives an estimate of how much an alpha particle could scatter: ~0.016 degrees S It assumes an elastic collision and uses conservation of energy and momentum: please go and read it S Since the gold foil consists of more than 1 electron, the effects of these scatters would add together S At the 2 nd scatter, the alpha could go further away from the horizontal axis or come back. One can show that <θ> = N * θ_single 4
Calculation S In the gold foil, of thickness = 6 x 10^-7 m, one can estimate that there are ~2,300 atoms S This is a straightforward estimation; using the density and molar mass or atomic weight of gold, one has S # molecules / cm^3 = 6 x 10^23 molecules/mole * (1 mol / 197 g) * 19.3 g/cm^3 = 5.9 x 10^28 atoms per m^3 S If they re spread uniformly, then each atom occupies (1/5.9x10^28) m^3 S Distance between centers is (1/5.9x10^28)^(1/3) m = 2.6 x 10^-10 m. S Therefore, in a 6 x 10^-7 m thick foil one would have ~ 6 x 10^-7 div. by 2.6 x 10^-10 ~ 2,300 atoms. So, one expected that the alpha particles would scatter by an S => <θ> = 2300 * 0.016 ~ 0.8 5 average of less than one degree!
A Surprise S Much to everyone s shock and surprise, they found that, at times, the alpha particles were scattering by as much as 90 or even by 180! S Someone remarked that it was as if you shot a cannon shell at a piece of tissue paper, and that metallic shell bounced back! S This observation was inconsistent with a Thomson model, and led Rutherford to propose the more familiar model of a nucleus (positive) and electrons circulating around it S The large scatters were the alpha s hitting the nucleus S The electrons had to be moving to avoid falling into the nucleus => very much like how planets move in orbits around the sun. 6
Rutherford Scattering S We won t derive formula, but it is illustrative to go over the arguments. S Idea was: the α particle encountered the electric field of the nucleus S The closer it came, the larger the force, hence larger the scattering angle. S With some simple assumptions, one can derive relationship for b & θ S M_nucleus >> M_alpha, so nucleus recoil is neglected. S Thin target =>assume only nuclear scatter S Assume that the alpha and the nucleus are point(-like) particles 7 Only the Coulomb force is in play, also
Newton S One essentially uses Newton s law, F=ma or F = dp/dt (p = mv) => p = F * dt S Since the alpha-particle changes direction => dp/dt not equal to 0 S This change is due to the Coulomb force ( = Z1*Z2*e^2 / (4*π*ε 0 *r^2) * r^, where r^ or r-hat is the direction of the force). S In reality, life is a bit more complicated: since the α - particle feels the electric field from Z 2 *e continuously, then one has to integrate to account for it S At different times, the direction of F will be different S See full derivation on page 133 8
Final Result S All this gives us b = [ Z 1 Z 2 e^2 / (8 π ε 0 KE) ] cot ( θ / 2 ) S Where KE is (1/2)*m*v 0^2, the incident kinetic energy S Remember, we are shooting a beam of alpha-particles at a foil, and we don t know the locations of atoms, so we do not really know b for each collision S Also, no experiment perfect, so our measurement of θ will have uncertainty S Rest of the derivation deals with how to account for this -> essentially, we measure the fraction of scattered particles at various angles 9
The Classical Atomic Model S Example 4.3 shows you how to use this formula. For the values given, we get that the fraction of scatters at 45 is ~10^-7 per mm^2 S It is a very small fraction, but non-zero. S Section 4.3: Rutherford s experiments led scientists to believe that the atom consisted of a heavy nucleus surrounded by electrons. S They would have to be moving; otherwise, they d fall into the nucleus. S However, moving electrons should give off EM waves => they will continuously lose power S And very quickly fall into the nucleus (~10^-9 s). Oops! 10
The Bohr Model S So, we have a problem! Niels Bohr came up with a solution. It has some interesting assumptions S Atoms have stationary states, i.e., states where electrons do not radiate energy while circulating around the nucleus -> definite energy. S EM radiation (e.g., line spectra) occur when electrons move from one stationary state to another. E of radiation = h*f = E1 - E2. S When moving in stationary orbits, laws of classical physics are valid, but not during transitions S Mean KE of electron-nucleus system 11
Section 4.4 of R&T S With these assumptions, Bohr was able to predict the Rydberg formula of line spectra S See the relevant pages of lecture notes for Chapter 3 S Total E of electron-nucleus system: S Assume M n -> infinity, so it does not move. S Total E = KE elec + V, where V is the potential energy S = (1/2) m v^2 e^2 / ( 4 π ε 0 r ) S As the electron revolves around the nucleus, it has a = v^2 / r = e^2 / ( 4 π ε 0 r^2 ) * ( 1 / m ) the electric force => v^2 = e^2 / ( 4 π ε 0 r m ) S Total E = (m/2)e^2/( 4 π ε 0 r m ) e^2/( 4 π ε 0 r ) = (the negative sign implies bound system) 12 (centripetal)
Energy S E n = -13.6 ev / n^2 S Implies that for n = 1, E = -13.6 ev -> electron is most tightly bound. S As n increases, E n decreases S As it turns out, it takes ~13.6 ev to strip out the e- from a H atom!! S When an electron is in, say, the state n = 2 its E = -13.6 / 2^2 S If it now transitions to state 1, then E of emitted photon = h*f = +13.6*(1-1/2^2) S For transitions between any two states n_upper and n_lower we get h*f = 13.6 * (1/n_lower^2 1/n_upper^2) S Or, because λ=c/f -> 1/λ=f/c =>1/λ = (13.6/hc) (1/n l^2 1/n u^2) 13
Blast from the Past S Recall that in Chapter 3 (page on Rydberg) we derived that S 1 / lambda = R H * ( 1 /n^2 1 / K^2 ), where R H is nothing but 13.6/ (hc) and we can identify n and K with *n_lower and n_upper* S So, Bohr s model was able to explain the line spectra! S However, it too had problems (and is actually not totally correct) S An aside on the board: the derivation of the fine structure constant, which appears everywhere in quantum mechanics 14
Bohr s Correspondence Principle S Classical and quantum domains are different S Planck s constant helps with deciding where we are: e.g., energy of a baseball is classical while the energy of a moving electron is quantum S h = 6.626 x 10^-34 J-s -> E of baseball =(1/2)mv^2 is ~Joules and m ~ 0.1 kg while v ~ 90 mph ~ 40 m/s so E ~ 0.5 * 0.1 * 1600 ~ 80 J >> h / t S As the mass m of a baseball decreases to the electron mass value, then we suddenly find ourselves in the quantum domain S The correspondence principle stated that as we go from the quantum domain to the classical domain, then the quantum result should yield the classical result. (Of course it is continuous) 15
Problems with the Bohr Model S It has some unjustified assumptions S It only worked for single e- atoms S Could not explain binding of atoms into molecules S Had trouble with precision line spectrum data S We will return to this after introducing true Quantum Mechanics S No homework for now. Next assignment will cover both Chapters 4 & 5 16