Physics 9 Fall 2011
Homework 11 - Solutions
Wednesday November 30, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t
finish them. The homework is due at the beginning of class on Wednesday, December
7th. Because the solutions will be posted immediately after class, no late homeworks can
be accepted! You are welcome to ask questions during the discussion session or during office
hours.
1. Two narrow slits are separated by a distance d. Their interference pattern is to be
observed on a screen a large distance L away.
(a) Calculate the spacing between successive maxima near the center fringe for light
that has a 500 nm wavelength, when L is 1.00 m and d is 1.00 cm.
(b) How close together should the slits be placed for the maxima to be separated by
1.00 mm for this wavelength and screen distance?
————————————————————————————————————
Solution
The double-slit diffraction equation is
d sin θm = mλ,
where θm is the angle through which the mth order is diffracted, and λ is the wavelength. Now, since d λ then θ 1 (in radians). So, we can replace sin θ by θ.
Hence, for small angles
dθm = mλ.
Now, if the light is bent through an angle θm and impacts on a wall a distance L away,
then it is diffracted through a distance
ym = L tan θm .
Again, though, if θm is small, then tan θm ≈ θm , and so
ym ≈ Lθm = m
λL
,
d
where we have plugged in the diffraction equation. Now we can proceed with our
problem.
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(a) The spacing between successive spacings is
∆y = ym+1 − ym = (m + 1)
λL
λL
λL
−m
=
,
d
d
d
which doesn’t depend on the order (for these small angles). Plugging in the
numbers we find
∆y = ym+1 − ym =
λL
500 × 10−9 × 1
=
= 5 × 10−5 m = 50 µm.
d
10−2
(b) Now, solving for d gives
d=
500 × 10−9 × 1
λL
=
= 5 × 10−4 m = 0.5 mm.
∆y
10−3
2
2. The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength
in air equal to 400 nm and 1.61 for red light that has a wavelength in air equal to 700
nm. A ray of 700 nm wavelength red light and a ray of 400 nm wavelength violet light
both have angles of refraction equal to 30◦ upon entering the glass from air.
(a) Which is greater, the angle of incidence of the ray of red light or the angle of
incidence of the ray of violet light? Explain your answer.
(b) What is the difference between the angles of incidence of the two rays?
————————————————————————————————————
Solution
Recall Snell’s law for refraction,
n1 sin θ1 = n2 sin θ2 ,
where n1 is the index of refraction of the first material, θ1 is the angle of incidence, n2
is the index of refraction of the second material, and θ2 is the angle of refraction.
(a) Solving Snell’s law for the angle of incidence gives
−1 n2
θ1 = sin
sin θ2 .
n1
Because the index of refraction for violet light is larger than that of red light,
and since both rays have the same angle of refraction, θ1 for violet light is bigger
than that for red light. Thus, to exhibit the same refraction angle, the violet light
would need an angle of incidence larger than that of red light.
(b) The difference in the angles of incidence is ∆θ = θv − θr , where θv is the angle for
violet light, and θr is the angle for red light. From above, then (taking n1 = 1 for
air)
∆θ = sin−1 [nv sin θ2 ] − sin−1 [nr sin θ2 ] ,
where nv is the index for violet light, and nr is the index for red light. Furthermore,
taking θ2 = 30◦ , gives
h i
h i
−1 nr
−1 1.66
−1 1.61
−1 nv
− sin
= sin
− sin
= 2.49◦ .
∆θ = sin
2
2
2
2
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3. A laser beam is fired through the air
(with index nair = 1) at a plate of
clear material with thickness t and
index n, as seen in the figure to the
left. The beam initially makes an
angle of θ1 with respect to the normal of the material.
(a) Show that the beam exits the material along the same direction as it entered (i.e.,
show that the angle it makes with the normal upon exiting at the bottom is the
same angle that it entered with, θ1 ).
(b) The laser beam exits displaced to its initial path by a distance d, as seen in
the figure. What is this distance in terms of θ1 , θ2 , and the thickness, t, of the
material?
————————————————————————————————————
Solution
(a) The laser refracts through an angle θ2 upon entering the material, as given by
Snell’s law. Setting nair = 1, then sin θ1 = n sin θ2 .
Now, the laser travels along inside the material
along the path seen, until it reaches the bottom edge of the material. In this case, it makes
an angle θ3 with respect to the normal at the
bottom. Although it’s obvious that θ2 = θ3
from the diagram to the left, we can prove this
using a little bit of geometry.
θ1
θ2
θ3
θ4
The sum of the angles of the θ2 triangle have to be 180◦ , and since the angle on the
left is 90◦ , we see that the remaining angle to the right has to be 180◦ − 90◦ − θ2 .
This angle, plus θ3 make a right angle, and so θ3 +(180◦ − 90◦ − θ2 ) = 90◦ . Solving
gives θ3 = θ2 , as claimed. When it exits the material, the beam makes an angle
θ4 , which is again given by Snells law as n sin θ3 = sin θ4 . But, since θ3 = θ2 ,
n sin3 = n sin2 = sin θ4 . But, we’ve already seen that n sin θ2 = sin θ1 . Thus, we
see that sin θ1 = sin θ4 , or θ1 = θ4 . So, it does travel along the same direction.
(b) From the figure given in the problem, we see that the distance d = l sin φ =
l sin (θ1 − θ2 ). But, we want this in terms of the thickness, t. Again, from the
figure, t = l cos θ2 , and so
sin (θ2 − θ1 )
d = l sin (θ2 − θ1 ) =
t.
cos θ2
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4. Electrons with a speed of 2.0 × 106 m/s pass through a double-slit apparatus. Interference fringes are detected with a fringe spacing of 1.5 mm.
(a) What will the fringe spacing be if the electrons are replaced by neutrons with the
same speed?
(b) What speed must the neutrons have to produce interference fringes with a spacing
of 1.5 mm?
Hint: you can assume that the diffraction angle is small for this problem.
————————————————————————————————————
Solution
The electrons obey the usual diffraction equation,
d sin θm = mλ,
where the wavelength of the electron is the de Broglie wavelength,
λ=
h
.
p
Thus,
d sin θm =
mh
.
p
Now, for small angles sin θm ≈ θm , and so the position of the fringes on a screen a
distance L away is ym = L tan θm ≈ Lθm ⇒ ym /L. Thus, we can write
ym = m
hL
.
dp
Finally, the spacing between two adjacent fringes (for small angles) is
∆y = ym+1 − ym = (m + 1)
hL
hL
hL
−m
=
.
dp
dp
dp
Now we’re ready to approach the problem.
(a) Because the neutron is about 1840 times heavier than the electron, then we can
write
hL
1
hL
∆yelectron
∆yneutron =
=
=
.
dpneutron
1840 dpelectron
1840
Now, ∆yelectron = 1.5 mm, and so
∆yneutron =
1.5 × 10−3
∆yelectron
=
= 8.15 × 10−7 meters,
1840
1840
or about 820 nanometers.
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(b) If we want the fringe spacing to be the same as for the electrons, then we need
∆yneutron =
hL
dpneutron
=
hL
dpelectron
= ∆yelectron ⇒ pneutron = pelectron .
In other words, the neutron and electron have to have the same momentum. Since
the neutron is 1840 times heavier than the electron, it must also travel 1840 times
slower than the electron to have the same momentum. Thus,
vneutron =
velectron
≈ 1100 m/s.
1840
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5. (a) What are the energies of the first three energy levels of an electron confined in a
one-dimensional box of length 0.70 nm? Give your answer in electron volts (eV).
(b) How much energy must the electron lose to move from the n = 2 energy level to
the n = 1 energy level? Again, give your answers in eV.
(c) Suppose that an electron can move from the n = 2 level to the n = 1 level
by emitting a photon of light. If energy is conserved, what must the photon’s
wavelength be? Give your answer in nanometers.
————————————————————————————————————
Solution
(a) The energy of an electron in a box of length L is
En =
h2 2
n.
8mL2
For a box of length 0.70 nm, then
2
En =
(6.63 × 10−34 )
h2 2
n
=
n2 = 1.23 × 10−19 n2 = 0.77n2 eV.
2
2
8mL
8 (9.11 × 10−31 ) (.70 × 10−9 )
Thus, the energies for the n = 1, n = 2, and n = 2 states are, respectively,
E1 = 0.77 (1)2 eV = 0.77 eV
E2 = 0.77 (2)2 eV = 3.07 eV
E3 = 0.77 (3)2 eV = 6.92 eV.
(b) To move from the n = 2 level to the n = 1 level, the electron must lose an energy
E2 − E1 = 3.07 − 0.77 = 2.3 eV.
(c) A photon emitted in this process carries off the excess energy. The energy of a
photon is given in terms of its frequency as E = hf , but since λf = c, then we
can write E = hc/λ. So, solving for the wavelength we have
λ=
hc
.
E
Now, the energy emitted by the transition is found from part (b) as 2.3 eV. Now,
it turns out that, by making unit conversions, we can write hc = 1240 eV · nm,
which is very convenient for this calculation. Thus,
λ=
hc
1240 eV · nm
=
= 539 nm.
E
2.3 eV
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6. The Failure of Classical Mechanics! This problem is extra credit!
As we have discussed, the classical model of an atom, where the electron orbits the
proton at an arbitrary radius, is unstable. As the electron orbits the proton, it accelerates, and in accelerating it radiates energy. According to the Larmor formula, the
power, P = dE
, that is being radiated away as the electron accelerates is
dt
dE
2 ke2
= − 3 a2 ,
dt
3 c
1
where k = 4π
is Coulomb’s constant, e is the charge on an electron, and a is the
0
acceleration. The power is negative because the energy is being radiated away. What
we want to do is to estimate how much time it will take for the electron to spiral in to
the nucleus of the hydrogen atom.
(a) Using Coulomb’s law,
q show that the velocity of the electron at a distance r around
the nucleus is v =
ke2
.
mr
(b) Using your answer above, show that the total energy (kinetic plus potential) of
2
.
the hydrogen atom is E = − ke
2r
(c) Take the time derivative of E and equate your expression to the Larmor formula
dr
= dE
. You should find that
above. Hint: from the chain rule, dE
dt
dr dt
dr
4 r 2 a2
=−
.
dt
3 c3
(d) The electron is held to the proton by the electrostatic force, which Newton tells
us is also ma. Using this, show that your result from part (c) can be written as
dr
4 k 2 e4 1
=−
.
dt
3 m2 c3 r2
(e) Finally, integrate your result to determine how long it would take the electron,
starting at a radius r0 at time t = 0 to spiral down to the nucleus, which we’ll take
to be zero radius. If the electron starts at the classical Bohr radius, r0 = 0.528 Å,
what is the actual time in seconds?
————————————————————————————————————
Solution
(a) Since the electron is going around in a circle, then the force on it is mv 2 /r. But,
it’s also being held in its orbit by the electrostatic force, FEp= ke2 /r2 . Setting
these two forces equal and solving for the velocity gives v = ke2 /mr.
(b) The total energy of the atom is its kinetic energy ( 21 mv 2 ) plus it’s potential energy,
which for two oppositely-charged point particles is just −ke2 /r. So, the total
energy is
2
1 2 ke2
1
ke
ke2
ke2
E = mv −
−
= m
=−
,
2
r
2
mr
r
2r
where we have plugged in our result from part (a) for the velocity.
8
2
2
dr
dr
(c) Since E = − ke
, then dE
= dE
= ke
. Setting this equal to the Larmor for2r
dt
dr dt
2r2 dt
ke2 dr
2 ke2 2
mula gives 2r2 dt = − 3 c3 a . Canceling off the common factors gives the expected
results.
2
= ma, where m is the
(d) The electrostatic force is just Coulomb’s law, F = ke
r2
ke2
mass of the electron, and so a = mr2 . Plugging in this result to part (c) gives
2 2
4 r2 a2
ke
k2 e4 1
4 r2
dr
=
−
=
−
= − 43 m
2 c3 r 2 , as advertised.
dt
3 c3
3 c3
mr2
Rt
2 c3 R 0
2
r
dr
=
dt, which gives
(e) Here we just separate and integrate, − 34 m
2
4
k e
0
r0
t=
m2 c3 r03
.
4k 2 e4
Taking r0 = 0.528 Å, and plugging in numbers gives t ≈ 1.55 × 10−11 seconds!
Clearly you can see why we need quantum mechanics!
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