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 θ m
= mλ, where θ m is the angle through which the m th 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 dθ m
= mλ.
Now, if the light is bent through an angle θ m then it is diffracted through a distance and impacts on a wall a distance L away, y m
= L tan θ m
.
Again, though, if θ m is small, then tan θ m
≈ θ m
, and so y m
≈ Lθ m
= m
λL d
, where we have plugged in the diffraction equation. Now we can proceed with our problem.
1

(a) The spacing between successive spacings is
∆ y = y m +1
− y m
= ( m + 1)
λL d
− m
λL d
=
λL
, d which doesn’t depend on the order (for these small angles). Plugging in the numbers we find
∆ y = y m +1
− y m
=
λL d
=
500 × 10
− 9 × 1
10 − 2
= 5 × 10
− 5 m = 50 µ m .
(b) Now, solving for d gives d =
λL
∆ y
=
500 × 10
− 9 × 1
10 − 3
= 5 × 10
− 4 m = 0 .
5 mm .
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, n
1 sin θ
1
= n
2 sin θ
2
, where n
1 is the index of refraction of the first material, θ
1 is the angle of incidence, is the index of refraction of the second material, and θ
2 is the angle of refraction.
n
2
(a) Solving Snell’s law for the angle of incidence gives
θ
1
= sin
− 1 n
2 sin θ
2
.
n
1
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, θ
1 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 = θ v
− θ r
, where θ v is the angle for violet light, and θ r is the angle for red light. From above, then (taking n
1
= 1 for air)
∆ θ = sin
− 1
[ n v sin θ
2
] − sin
− 1
[ n r sin θ
2
] , where n v taking θ
2 is the index for violet light, and n r
= 30
◦
, gives is the index for red light. Furthermore,
∆ θ = sin
− 1 h n v i
2
− sin
− 1 h n r i
2
= sin
− 1
1 .
66
2
− sin
− 1
1 .
61
2
= 2 .
49
◦
.
3

3. A laser beam is fired through the air
(with index n air
= 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 θ
1 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, θ
1
).
(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 θ
1
, θ
2
, and the thickness, t , of the material?
————————————————————————————————————
Solution
(a) The laser refracts through an angle θ
2
Snell’s law. Setting n air
= 1, then sin θ
1 upon entering the material, as given by
= n sin θ
2
.
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 θ
3 with respect to the normal at the bottom. Although it’s obvious that θ
2
= θ
3 from the diagram to the left, we can prove this using a little bit of geometry.
θ
1
θ
2
θ
3
θ
4
The sum of the angles of the left is 90
◦
θ
2 triangle have to be 180
◦
, and since the angle on the
, we see that the remaining angle to the right has to be 180
◦ − 90
◦ − θ
2
.
This angle, plus θ
3 make a right angle, and so θ
3
+(180
◦ − 90
◦ − θ
2
) = 90
◦
. Solving gives
θ
4
θ
3
= θ
2
, as claimed. When it exits the material, the beam makes an angle
, which is again given by Snells law as n sin θ
3
= sin θ
4
. But, since θ
3
= θ
2
, n sin
3
= n sin
2 see that sin θ
1
= sin θ
4
. But, we’ve already seen that n sin θ
= sin θ
4
, or θ
1
= θ
4
2
= sin θ
1
. Thus, we
. 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 ( θ
1
− θ
2
). But, we want this in terms of the thickness, figure, t = l cos θ
2
, and so t . Again, from the d = l sin ( θ
2
− θ
1
) = sin ( θ
2
− θ
1
) cos θ
2 t.
4

4. Electrons with a speed of 2 .
0 × 10 6 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 θ m
= mλ, where the wavelength of the electron is the de Broglie wavelength,
λ = h p
.
Thus, d sin θ m
= mh
.
p
Now, for small angles sin distance L away is y m
= L
θ m tan
≈ θ m
, and so the position of the fringes on a screen a
θ m
≈ Lθ m
⇒ y m
/L . Thus, we can write y m
= m hL dp
.
Finally, the spacing between two adjacent fringes (for small angles) is
∆ y = y m +1
− y m
= ( m + 1) hL dp
− m hL dp
= hL
.
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
∆ y neutron
= dp hL neutron
=
1
1840 dp hL electron
=
∆ y electron
.
1840
Now, ∆ y electron
= 1 .
5 mm, and so
∆ y neutron
=
∆ y electron
1840
=
1 .
5 × 10
− 3
1840
= 8 .
15 × 10
− 7 meters , or about 820 nanometers.
5

(b) If we want the fringe spacing to be the same as for the electrons, then we need hL
∆ y neutron
= dp neutron hL
= dp electron
= ∆ y electron
⇒ p neutron
= p electron
.
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, v neutron
= v electron
1840
≈ 1100 m/s .
6

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
E n
= h 2
8 mL 2 n
2
.
For a box of length 0 .
70 nm, then
E n
= h 2
8 mL 2 n
2
=
(6 .
63 × 10
− 34 )
2
8 (9 .
11 × 10 − 31 ) ( .
70 × 10 − 9 )
2 n
2
= 1 .
23 × 10
− 19 n
2
= 0 .
77 n
2 eV .
Thus, the energies for the n = 1, n = 2, and n = 2 states are, respectively,
E
1
= 0 .
77 (1)
2
E
2
= 0 .
77 (2)
2
E
3
= 0 .
77 (3)
2 eV = 0 .
77 eV eV = 3 .
07 eV eV = 6 .
92 eV .
(b) To move from the n = 2 level to the n = 1 level, the electron must lose an energy
E
2
− E
1
= 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
=
E
1240 eV · nm
2 .
3 eV
= 539 nm .
7

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 dt
, that is being radiated away as the electron accelerates is dE dt
= −
2
3 ke 2 a
2
, c 3 where k =
1 is Coulomb’s constant, e is the charge on an electron, and a is the
4 π
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, show that the velocity of the electron at a distance r around the nucleus is v = q ke 2 mr
.
(b) Using your answer above, show that the total energy (kinetic plus potential) of the hydrogen atom is E = − ke
2
2 r
.
(c) Take the time derivative of E and equate your expression to the Larmor formula above.
Hint: from the chain rule, dE dt
= dE dr dr dt
.
You should find that dr dt
= −
4
3 r 2 a 2
.
c 3
(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 dt
= −
4
3 k
2 e
4 m 2 c 3
1 r 2
.
(e) Finally, integrate your result to determine how long it would take the electron, starting at a radius r
0 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, r
0 what is the actual time in seconds?
= 0 .
528 ˚ ,
————————————————————————————————————
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, F
E
= ke
2
/r
2
. Setting these two forces equal and solving for the velocity gives v = p ke 2 /mr .
(b) The total energy of the atom is its kinetic energy (
1 mv 2 ) plus it’s potential energy,
2 which for two oppositely-charged point particles is just − ke 2 /r . So, the total energy is
E =
1
2 mv
2 − ke
2 r
=
1
2 m ke
2 mr
− ke
2 r
= − ke
2
2 r
, where we have plugged in our result from part (a) for the velocity.
8

(c) Since E = − mula gives ke
2 ke
2
2 r dr dt results.
2 r
2
, then
= − 2
3 dE dt ke
2 c
3
= dE dr dr dt
= ke
2
2 r
2 dr dt
. Setting this equal to the Larmor fora 2 . Canceling off the common factors gives the expected
(d) The electrostatic force is just Coulomb’s law, mass of the electron, and so dr dt
= − 4
3 r
2 a
2 c 3
= − 4
3 r
2 c 3 ke
2 mr 2
2 a =
= − 4
3 k
2 m 2 e
4 c 3
1 r 2
F = ke
2 r 2
, as advertised.
= ma , where m is the ke
2 mr 2
. Plugging in this result to part (c) gives
(e) Here we just separate and integrate, − 3
4 m k 2
2 c
3 e 4
R
0 r
0 r 2 dr =
R t dt , which gives
0 t = m
2 c
3 r
3
0
.
4 k 2 e 4
Taking r
0
= 0 .
528 ˚ , and plugging in numbers gives
Clearly you can see why we need quantum mechanics!
t ≈ 1 .
55 × 10
− 11 seconds!
9