Assignment 4, solutions
Problem 1:
(a) Compare the intensity of a light bulb
at a distance of 6 m from it to the intensity at 2 m from it.
Repeat this comparison for laser light.
Explain fully but briefly.
(b) Compute
the electric field corresponding to a focused light intensity of I = 1012W/cm2,
and compare the result with the electric field experienced by the electron in a
Hydrogen atom.
Solution:
Problem 2:
Electrons in a computer monitor CRT are accelerated to a final
kinetic energy of 30 keV over a distance of 1 cm, then are rapidly
decelerated to zero speed in collisions with the screen phosphor.
Assume both acceleration and deceleration are constant. Consider
the energy radiated by accelerated electrons (which has nothing directly
to do with the light emitted by the phosphor).
(a) Can this problem be treated non-relativistically? Explain
why or why not.
(b) Develop an expression for the ratio r of the energy
radiated during the acceleration phase, Erad, to the
final kinetic energy Ekin, assuming constant acceleration a.
Also calculate a numeric value for r under the conditions
pertaining to the acceleration of electrons in the monitor CRT described
above.
(c) Again assuming constant acceleration, estimate the maximum
total fraction of kinetic energy that is radiated during the stopping of
the electrons in the phosphor, and from that, the average power radiated
per stopped electron in watts. Assume all the kinetic
energy is consumed in single collisions in a distance of 0.05 nm within
single atoms of the phosphor.
Solution:
 | Concepts, principles, relations
that apply to the problem:
Radiation produced by accelerating charges, the Lamor formula |
| Why do they apply?
We are asked to determine the energy radiated by an accelerating electron. |
| How do they apply?
(a)
The rest energy of the electron is 510 keV, its maximum kinetic energy is 30
keV.
gmax510 = 540,
gmax = 1.06 @ 1. The
problem can be treated nonrelativistically.
(b) An accelerating charged particle radiates away energy. If for
a time T the magnitude of the acceleration is a, then the energy radiated Erad
= 2e2a2T/(3c3) is given by the Lamor formula.
If the particle is initially at rest, and we neglect radiation, then the
kinetic energy of the particle after time T will be Ekin = (1/2)ma2T2.
Erad << Ekin, if 2e2a2T/(3c3)
>> (1/2)ma2T2, T >> (4/3)e2/(mc3)
= (4/3)r0/c.
For an electron r0
@ 3*10-15m.
If T is much greater than the time it takes light to travel 4*10-15m
then we can neglect radiation losses when considering the short term motion.
Erad/Ekin = 4e2/(3c3mT). |
|
Details of the calculation:
(b) (1/2)ma2T2 = 30000*1.6*10-19J.
a = (30000V/0.01m)(1.6*10-19C/9.1*10-31kg).
T = 1.94*10-10s. Erad/Ekin = 6.4*10-14.
(c) Ekin = (1/2)ma2t2 = 30 keV.
From kinematics: d = (1/2)at2. Therefore a = Ekin/md.
t2 = 2d2m/Ekin = 2(5*10-11m)2(9.1*10-31kg)/(3*104*1.6*10-19J).
t = 9.7*10-19s.
Erad/Ekin = 1.3*10-5. <P> = Erad/t
= 0.06W. |
Problem 3:
(a) An electron orbits initially, at time t = 0 around a
proton at a radius a0 equal to the Bohr radius. Using classical
mechanics and classical electromagnetism derive an expression for the time it
takes for the radius of the orbiting electron to decrease to zero due to
radiation. Here you may assume that the energy loss per revolution is small
compared to the total energy of the atom.
(b) What implication can you draw from this calculation?
Give a qualitative argument on the need to modify the above estimate.
Lamor formula:
P = -dE/dt = [q2/(6pe0c3)]a2.
(SI units)
Solution:
| Concepts, principles, relations that apply to the
problem:
The
radiation field of a point charge moving non-relativistically, the Lamor
formula |
| Why do they apply?
The electron
orbits the "infinitely heavy" proton. Its acceleration is a = v2/r,
and it therefore radiates away energy,
P = -dE/dt = [q2/(6pe0c3)]a2.
(SI units) |
| How do they apply?
(a) To solve for the
total time it takes the electron to spiral into the nucleus, we can express the
acceleration a in
terms of E. We then have
-dE/dt = f(E), òdt
= Dt = òE0-¥dE/f(E).
Alternatively, we can express E as a function of the radius r and derive an
expression -dr/dt = f(r). We then have
òdt =
DT = òr00dr/f(r).
F = q2/(4pe0r2) = mv2/r. mv2 = q2/(4pe0r).
E = (1/2)mv2 - q2/(4pe0r)
= - q2/(8pe0r) = -(1/2)mv2.
a = v2/r = q2/(4pe0r2m)
= 2E28pe0/(q2m).
-dE/dt = E4162pe0/(6c3q2m2).
dt = -(dE/E4) (6c3q2m2)/(162pe0).
Dt
= (6c3q2m2)/(162pe0)òE0-¥(dE/E4) = -(6c3q2m2)/(162pe0)(E0-3/3)
where E0 = -13.6 eV,
Dt
= 1.5*10-11s.
or
a = F/m = q2/(4pe0r2m),
E = - q2/(8pe0r),
dE = [q2/(8pe0r2)]dr
= -Pdt.
dt = -[q2/(8pe0r2)]dr/P
= -[q2/(8pe0r2)]dr/[q2a2/(6pe0c3)].
dt = -dr[3c3/(4r2a2)] = -r2dr[12c3p2e02m2/q4].
Dt
= -[12c3p2e02m2/q4]òr00r2dr
= [12c3p2e02m2/q4](r03/3)
where r0 = Bohr radius = 5.29*10-11m,
Dt = 1.5*10-11s.
(b) This result
implies that the atom is unstable. Since atoms exist, the classical theory
needs to be modified. |
| Details of the calculation:
None |
Problem 4:
Consider a linear antenna of length d (d << l)
with a narrow gap in the center for the purposes of excitation. Assume
that the current is sinusoidal and in the same direction in each half of the
antenna, having a value of I0 at the gap and falling linearly to zero
at the ends. Find the power radiated in the electric dipole approximation.
Solution:
| Concepts, principles, relations
that apply to the problem:
Radiation from antennas, dipole radiation |
| Why do they apply?
We can treat the antenna as made up of small sections, each section emitting
dipole radiation. |
| How do they apply?
Consider a small section of the antenna of length dz' located at z' . Treat it like a dipole.
Let p(z') = qdz'(z/z), q(z') = q0(z')coswt.
Then I(z') = dq(z')/dt = -wq0(z')sinwt
= I0(z')sin(wt), I0(z') = -wq0(z').
p(z') = p0(z')coswt. p0(z') = q0(z')dz
= -(I0(z')/w)dz.
For dipole at the origin we have ER(r,t) = -(1/(4pe0c2r))w2p0
cos(w(t-r/c))sinq .
A small section of the antenna of length dz' located at z' a distance R
= r - (z/z)z' from the observation point therefore produces
dER(r,t) = (1/(4pe0c2R))w2(I0(z')/w)cos(w(t-R/c))sinq'dz''.
dBR(r,t) = (1/(4pe0c3R))w2(I0(z')/w)cos(w(t-R/c))sinq'dz' .
Assume r >> l, r >> z', d <<
l.
Then q' @
q = constant, for all sections of the antenna.
We can also replace 1/R by 1/r and
cos(wt - kR) by cos(wt
- kr) because the changes in kR are small compared to p/2
over the length of the antenna.
I0(z) = I0 - (2I0/d)|z|, the current falls linearly to
zero at the ends.
Eq(r,t) = òz=-d/2z=d/2dEq(r,t)
= 2 ò0z=d/2dEq(r,t)
= (I0w/(2pe0c2r))cos(w(t-r/c))sinq
ò0z=d/2(1-2z/d)dz'
= (I0dw2/(8pe0c3))(1/(kr))cos(w(t-r/c))sinq.
Bf(r,t) = òz=-d/2z=d/2dBf(r,t)
= 2 ò0z=d/2dBf(r,t)
= (m0I0dw2/(8pc2))(1/(kr))cos(w(t-r/c))sinq. |
| Details of the calculation:
S = (1/m0)E´B
=  [
(I02d2w4/(64p2e0c5k2r2))cos2(w(t-r/c))sin2q.]
P = (I02d2w2/(64p2e0c3r2))cos2(w(t-r/c))
ò2pr2 sin3q
dq
= (I02d2w2/(32pe0c3))cos2(w(t-r/c))(4/3).
<P> = (1/2)(I02d2w2/(24pe0c3))
is the average power radiated in the electric dipole approximation. |
Problem 5:
Starting with Maxwell’s Equations:
(a) Derive the wave equations for a light wave in vacuum. Write out solutions
for these equations for E and B.
(b) Show that the electric and magnetic fields are in phase, perpendicular to
each other and perpendicular to the direction of motion.
(c) Determine the relative magnitude of the E and B fields.
Solution:
| Concepts, principles, relations that apply to the
problem:
Maxwell's equations |
| Why do they apply?
In regions where r and j are zero Maxwell's equations lead to the
homogeneous wave equation for E and B. All solutions can
be viewed as linear superpositions of sinusoidal plane wave solutions.
Inserting these solutions into Maxwell's equations we derive (b) and (c). |
| How do they apply?
(a) Maxwells equations in SI units are
 ,
 .
Assume r and j
are zero in the medium.
Ñ´(Ñ´E)
= Ñ(Ñ×E)
- Ñ2E = -(¶/¶t)(Ñ´B)
= -m0e0¶2E/¶t2.
Ñ2E = -m0e0¶2E/¶t2.
Ñ´(Ñ´B)
= Ñ(Ñ×B)
- Ñ2B = m0e0(¶/¶t)(Ñ´E) = -m0e0¶2B/¶t2.
Ñ2B = -m0e0¶2B/¶t2.
m0e0
= 1/c2. |
| Details of the calculation:
(b) Each Cartesian
component of E and B satisfies the 3-dimensional, homogeneous
wave equation.
Sinusoidal plane wave solutions E(r,t) = E0
exp(i(ki×r -
wt)), B(r,t) = B0
exp(i(ki×r -
wt)) exist.
Ñ×E
= ¶Ex/¶x +
¶Ey/¶y +¶Ez/¶z
= ik×E = 0
Ñ×E = 0
requires that E×k = 0 for radiation
fields, i.e. that E is perpendicular to k.
Similarly, Ñ×B
= 0 requires that B×k = 0 for radiation
fields.
Ñ´E = -¶B/¶t
requires that ik´E = iwB,
i.e. B is perpendicular to E and k.
(c) B = (k/w)E = E/c. |
Problem 6:
A plane electromagnetic (EM)
wave is incident on a free particle of charge q and mass m. The EM wave causes
the particle to oscillate and hence to radiate. The interaction can be
considered as a scattering of EM radiation with cross section
sT = (power
radiated)/(incident flux).
Assume the interaction can be
treated non-relativistically.
Using Larmor’s radiation formula, show that
.
Evaluate
sT for an electron.
Solution:
| Concepts, principles, relations that apply to the
problem:
Radiation produced by accelerating charges, the Lamor formula |
| Why do they apply?
We are asked to determine the power radiated by an accelerating electron and
compare this power to the incident flux. |
| How do they apply?
Let E = Ek. The force on the electron is
F = -qeE = ma.
Let E = E0eiwt at
the position of the electron. Then d2z/dt2 = -(qeE0/m)
eiwt.
z = z0eiwt, -w2z0
= -(qeE0/m), z0 = (qeE0/(mw2)).
The electron oscillates with frequency w
and amplitude z0. The total power radiated by such an electron is
<P> = qe2w4z02/(12pe0c3)
= qe4E02/(12pe0c3m2).
The light energy incident per unit area per unit time is given by the magnitude
of the average Poynting vector.
<S> = (1/2)e0cE02.
Scattering cross section:
s = <P>/<S> = qe4/(6pe02c4m2)
= (1/(4pe0))2(8p/3)(qe4/(c4m2)).
For an electron: s
= (8p/3)r02, where
r0 = e2/(mc2) = 2.82*10-15 m.
s
= 6.65*10-29 m2. |
| Details of the calculation:
None |
Problem
7:
In a purely classical model we
consider a dielectric medium as a collection of uncoupled classical harmonic
oscillators. Assume that each oscillator consists of an electron connected to a
fixed ion by a harmonic spring with frequency w0.
(a) Write down and solve the equation of motion for the electron when a
monochromatic electric field with frequency w
is applied.
(b) For an electron density n, calculate the electric polarization and the
dielectric constant e(w).
(c) For a free electron gas, at what frequency is
e = 0? What is the physical
significance of this frequency?
Solution:
| Concepts, principles, relations that apply to the
problem:
The driven harmonic oscillator, polarization, the plasma frequency |
| Why do they apply?
We model a medium interacting with an electromagnetic as a collection of
driven harmonic oscillators. |
| How do they apply?
(a) Let E = E0
exp(-iwt).
The electric field and the spring exert a
force on he electron. The total force is
F
= -qe E – mw02r
= md2r/dt2.
To solve this differential
equation try a solution of the form
r
= r0 exp(-iwt).
This yields
r0 = -(qeE0/m)/(w02
- w2).
(b) The induced dipole moment
is p = -qer = (qe2E0/m)/(w02
- w2) exp(-iwt).
Polarization = dipole moment
per unit volume
P = np = n(qe2E/m)/(w02
- w2) =
e0ceE.
e(w)
= e0(1 +
ce)
= e0
+ n(qe2/m)/(w02
- w2).
(c) For a free electron
w0 = 0.
We then have
e(w) =
e0
- nqe2/(m w2)
= e0(1 -
nqe2/(e0m
w2)) =
e0(1 -
wp2/w2).
When
w = wp
the e(w)
= 0. wp = (nqe2/(e0m))1/2
is called the plasma frequency. The plasma is opaque to EM waves with
frequencies less than wp and
transparent to EM waves with frequencies greater than
wp. |
| Details of the calculation:
None |
