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Unformatted text preview: Chapter 13 Classical Model of the
Atom (Thomson’s Model) Assume that an atom is modelled by a single electron of mass m and charge
6 embedded in a uniformly charged sphere (the nucleus) of radius a With total charge —e (Fig. 13.1).
Fig. 13.1 From the Maxwell’s equation
VE247rp, (13.1) and Gauss’ theorem of vector calculus [(3.7)] it can be shown that the electro—
static force on the electron due to the nucleus is given by 2 F : ——— . (13.2) Exercise 13.1 Prove the above statement. Assume this is the only force acting on the electron, then the equation of
motion is ._ 627‘
Equivalently,
9
.. 2 __ 2 __ ed
7 +w0r — O, we 2 ma3 . (13.4) This is the equation of motion of an undamped simple harmonic oscillator, with
the general solution
r = A cos wot + B sinwot, (13.5) 59 60 CHAPTER 13. THE CLASSICAL ATOM where . O
A = 7(0) , B = 31—) . (13.6) we
The motion is obviously bound and periodic. In the absence of dissipative forces, the total energy 1 1
E=K+U= 5mi~2+§kr2, kswgm, (13.7) is a constant of motion, and by (13.5), can be written as E = gauges + B2) = g (mow + wig(+(0))2) . (138) Exercise 13.2 Prove the above equation. From classical electrodynamics, however, an accelerating electron emits ra
diation and therefore will lose energy. We will use the following result for the
time rate of energy loss of an accelerating charge: dE 262 ,,
E = _E§<T2>’ (139) where the angular brackets mean averaging over one cycle of the oscillatory
motion of the electron. From (13.5), ii = —Aw3 cos wot — ng sin wot . (13.10)
Hence
’r"2 : A2111?) cos2 wot + 321113 sin2 wot —— 2A  B 11161 sin wot cos wot . (13.11)
It follows that
”9 1 T ..9 010 3+8 ..9 (4161 9 '1
‘ E— dt”=— dt“=—A“ B“. 13.12
<r>T/0r27r0r2<+> <> Eq. (13.9) then yields 2 2 4 2 2 2
dE _ __E_EQ(A2+B2) : _( e “0) (lrnwg(A2_§—B2)> E —'yE, (13.13) 7i?— 3c3 2 377109 2
where 2 0 9
6‘01" 7 2 gm; (13.14) is the decay constant for E. We thus have E(t) = E(0)e—1t . (13.15) 61 We can model this energy loss by including a damping force Fd in addition
to the electrostatic force F of (13.2). This can be done as follows. Starting from m'f' + kr = Fd , (13.16)
we have
méir—i—krrder, (13.17)
which can be rewritten as
d 1 .9 1 ., dE _ 2e2 9
~ g n ' _ _ __ . __ —— “ , 1.].
dt <2m7‘ 21w") dt Fd 'r 363 (7“ ) (3 8) where the last equality follows from (13.9). Now i3 = —(7' . 7'4) — 7% . —. (1319) (F...>=g{<..g><%(...)>}. (13.20) In order to compare the average values in the RHS of the above equation, we
assume that the damping is small and write Thus 7 N Acos(w0t + qt) , (13.21)
d
E; N —.Aw0 sin(w0t + (1)) , (1322)
d2? 2
E N —.Awo cos(w0t + (Z5) , (13.23)
d3'r' 3 .
3—73—3— N Awo s1n(wot + ¢) . (13.24)
Hence
d . .. d A2 3 .
dih  7“) N E < :10 s1n{2(w0t + qﬁ)}> : Azwg cos{2(w0t + qﬁ)} , (13.25)
which implies
(5‘5” . %)> ~ 0. (13.26)
On the other hand,
 d3?" 2 4  2
7'  F N —.A wo sm (wot + qt) . (13.27) Thus <+E£§>~— 20. (13.28) 62 CHAPTER 13. THE CLASSICAL ATOJVI It follows from (13.20) that . 262 , d37‘
Fd.TN§E§TE{3_, (13.29)
and thus 9 3
26“ d 7’
F N —— ~—— . 13.30
d 3C3 dt3 ( ) The equation of motion of an electron in the classical (Thomson’s) model of the
atom including radiative effects can ﬁnally be written as m'; + war —— —— —— = 0. (13.31) We provide an approximate solution of this third—order equation as follows. Assume .
r = Aemt . (13.32) Then d3? . v ' .. 9 '
7‘ = zwAeWt , 7“ = —w‘AeWt , ——3
dt = —z’w3AeW . (13.33) Substituting the above three quantities into the equation of motion (13.31) we obtain
262 377103 Exercise 13.3 Verify the above equation. In the third term on the LHS of (13.34) we will approximate 3 3 by the
me constant 7 [c.f. (13.14)]. The thirddegree characteristic equation (13.34) then
appears as the quadratic equation —w2 + 133+ 333 = 0. (13.34) 262w2 L122 — i'yw — tag 2 0, (13.35) which has the solutions ‘ 42_ 2
w=WiV “’0 7 ~1w0+ﬂ 2 2 2 , (13.36) where in the last (approximate) equality we have assumed that 7 << can. The
approximate solution of (13.31) is then r(t) = 6—7t/2(Aem°t +Ze‘iw0t) . (13.37) In the above equation, the complex conjugate of A appears in the second term
on the RHS due to the requirement that 'r'(t) is real. 63 We will now calculate the power spectrum P(w) of the emitted radiation.
Assume that the electron is released from rest at time t = 07 so that radiation
begins to be emitted at this instant. Through Parseval’s theorem [(12.42)] and
(13.9) we can deﬁne P(w) by 00 262 DO 9
P w dw = — r‘dt. (13.38)
[m < > 363 0 Note that the lower time limit t = 0 is due to our assumption that the electron
is released from rest at t 2: 0, that is, i'(t) : 0 for t < 0. A straightforward
differentiation of (13.37) yields 7‘05)
7_.'~’ _1_. 1.2— 1
(5 m0) Aexp{ (2 two) 75} + (2 +sz) Aexp{ (2 +sz) t},
I t > O ,
0 , t g 0.
(13.39)
Exercise 13.4 Verify the above equation.
Let g(w) be the Fourier transform of Ht), that is,
m) = 2i / dwg(w)e_iwt , 9(a)) = / gamma“ . (13.4.0)
71' —oo —00
Using (13.39) in the second equation of (13.40), we obtain
°° 7 . 2 7 . .
g u) =/ dt —zw0 Aexp — ——zw0—zw t
< > 0 (2 ) i (2 ) }
°° "y . 2— "y . .
+/ dt —+7,w Aexp ~ —+qu —zwt
0 <2 0) { (2 0 ) } (1341)
1 _  2 1 2
= (2 two) A g (2 +zw0) A
g—i(w0+w) g+i(w0—w)
It is clear from the above equation that
9(a)) : g(—w) . (13.42) Exercise 13.5 Show that (13.42) also follows from the requirement that r(t)
is real. 64 CHAPTER 13. THE CLASSICAL ATOM From (13.40) we have / 7' dt: OOOdt/C: dwg(w)e_1wtii(t)
0 1 ”'00
:dw g(— Inf/000 dt e_i‘”t‘r‘( 1 ——2——/ —
:27T 277 00 2—1— :dwg(w )g/OO dtr(t)e Wt ix: dwg—(w w)=2—17—r(/:o dwg(w )()+/000 MEMO»)
:—17r(/°°<dwg— wﬂ>+/Odwg—<m gm) ﬁ/mwgwm (42.) (13. 43)
Since we are integrating only from O to 00 (in to), we can ignore the anti resonant term in the explicit expression for g(w) given by (13.41)7 that is, the
one proportional to 1/ {g— — i(w0 + w)}, and obtain im’t fa) 9 ‘7 ()() (% +i(w0 —w)) (% — “we — w)) I E mrzdtz 262
3C3 0 37rc3 (13.44) We assume that the radiative damping of the atom is weak, so that 'y / 2 << we.
In this case the above integrand is sharply peaked near w N we and the lower
limit of the integral can be replaced by —00. On comparison with (13.38), we
arrive at the following result for the power spectrum of a radiating (classical)
atom: (13.45) The spectral line given by this equation is said to have a Lorentzian lineshape. ...
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 Spring '08
 LAM
 mechanics

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