Download Chapter 4: Struct of Atom

The Structure of the
Atom
Chapter 4 in Rex and Thornton’s “Modern Physics”
Wed. October 26, 2016
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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 19th century
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Competing Models
S  The Thomson Model
aka the “plum-pudding” 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
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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 2nd scatter, the alpha could go further away from the horizontal
axis or come back. One can show that <θ> = N * √θ_single
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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
S  => <θ> = √2300 * 0.016 ~ 0.8°
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particles would scatter by an
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.
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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.
Only the
Coulomb
S  Thin target =>assume only nuclear scatter
force is in
S  Assume that the alpha and the nucleus are point(-like) particles play, also
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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*ε0*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 Z2*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
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Final Result
S  All this gives us b = [ Z1 Z2 e^2 / (8 π ε0 KE) ] cot ( θ / 2 )
S  Where KE is (1/2)*m*v0^2, 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
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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!
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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
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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 Mn -> infinity, so it does not move.
S  Total E = KEelec + V, where V is the potential energy
S  = (1/2) m v^2 – e^2 / ( 4 π ε0 r )
(centripetal)
S  As the electron revolves around the nucleus, it has a = v^2 / r = e^2 /
( 4 π ε0 r^2 ) * ( 1 / m ) the electric force => v^2 = e^2 / ( 4 π ε0 r m )
S  Total E = (m/2)e^2/( 4 π ε0 r m ) – e^2/( 4 π ε0 r ) =
(the negative sign implies bound system)
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Energy
S  En = -13.6 eV / n^2
S  Implies that for n = 1, E = -13.6 eV -> electron is most tightly bound.
S  As n increases, |En| 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/nl^2 – 1/nu^2)
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Blast from the Past
S  Recall that in Chapter 3 (page on Rydberg) we derived that
S  1 / lambda = RH * ( 1 /n^2 – 1 / K^2 ), where RH 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
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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)
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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
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