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Physics 9 Spring 2009
Final Exam - Part 2
For the final, you may use three sheets of notes with whatever you want to put on it, front and back. Please sit every other seat, and please don’t cheat! If something isn’t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have the full length of the final. If you attach any additional scratch work, then make sure that your name is on every sheet of your work. Good luck!
- Electrons accelerated through a potential of 40 kV pass through a thin metal foil made up of randomly oriented microcrystals and fall on a photographic plate 30 cm behind the foil.
(a) Calculate the de Broglie wavelength of the electron. (b) The innermost ring of the electron diffraction pattern has a diameter of 1. 7 cm. What is the spacing of planes of atoms in the microcrystals to which this ring corresponds?
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Solution
(a) The de Broglie wavelength is given by λ = hp , where p = mv is the momentum of the electron. The electron picks up speed falling through the potential. From conservation of energy, KE = eV , where e is the charge on the electron and V is the potential. Since KE = 12 mv^2 = p
2 2 m , then^ p^ =^
2 mKE =
2 meV. Thus, the de Broglie wavelength is
λ =
h √ 2 meV
6. 63 × 10 −^34
2 × 9. 11 × 10 −^31 × 1. 602 × 10 −^19 × 40 × 103
= 6. 14 × 10 −^12 m.
So, the de Broglie wavelength is about 0.0614 Angstroms (˚A). (b) The electrons diffract through an angle θ. From the geometry of the diffraction setup, tan θ = (^) Ly , where y = 0. 85 cm is the radius of the ring, and L = 30 cm is the distance away. So, θ = tan−^1
30
≈ 1. 62 ◦. Now, the diffraction equation says d sin θ = mλ. Solving for d gives d = (^) sinmλ θ. Since the ring is the innermost one, it is the first-order ring and so m = 1. We calculated λ from part (a), and so we just have to solve for d, using our result for θ. Thus,
d =
mλ sin θ
1 × 6. 14 × 10 −^12
sin (1.62)
≈ 2. 17 × 10 −^10 m
So, the atoms are spaced about 2.17 ˚A apart.
- The Large Hadron Collider (LHC) has been completed, and will restart very soon. The LHC is a circular particle accelerator with a circumference of 27 km, and accelerates protons up to a kinetic energy of 7 T eV , or 7× 1012 eV. Since they are moving at roughly 99 .9999991% of the speed of light, one needs Einstein’s theory of Special Relativity to properly calculate the momentum of the protons. However, if this is done, it is found that for particles moving this fast, the momentum, p = E/c, where E is the kinetic energy of the particle, and c is the speed of light. When they are injected, the protons are moving through a circular ring of radius R = 2803.98 meters (this is not the radius of the large ring, described above). Calculate the magnetic field, perpendicular to the motion of the proton, needed to keep the protons moving at this radius, with the momentum given above. ————————————————————————————————————
Solution
The magnetic force keeping the proton moving in a circle is Fmag = evB, where e is the charge on the proton, v is the velocity of the proton, and B is the magnetic field. Since the proton is moving in a circle, there is also a radial force on it F = mv
2 r. To have a stable orbit, at radius r = R, these two forces have to be equal. So,
evB =
mv^2 R
⇒ B =
mv eR
p eR
where p = mv is the particle’s momentum. Using this form, we can substitute for the momentum, p = E/c, and so B =
E
ecR
Now, we can just plug in the values and calculate things, but we’ll make one more simplification. Since E = 7 × 1012 eV , and 1 eV = 1. 602 × 10 −^19 J, we see that E/e = 7 × 1012 J/C. So, plugging in for c and R, we finally find
B =
E
ecR
7 × 1012
2. 99 × 108 × 2803. 98
= 8. 35 T.
Despite the approximations used, we find 8. 35 T. The actual value is 8. 33 T , and so we are very close!
- In 1965, Penzias and Wilson discovered the cos- mic microwave background (CMB) radiation left over from the Big Bang expansion of the Universe, which subsequently earned them the Nobel Prize in physics in 1978. This radiation background be- haves in a specific way, called a blackbody with a temperature of 2. 725 K, as seen in the figure. The temperature is extremely uniform, with fluctua- tions less than one part in 100,000! Precise mea- surements of the CMB are critical to cosmology, since any proposed model of the Universe must explain this radiation. These fluctuations in the temperature are thought to have formed the seeds of galaxy formation in the early Universe, as the CMB dates to roughly 379,000 years after the Big Bang. (a) The energy density of a blackbody (such as the CMB) is given by the expression
u =
π^2 k^4 B T 4 15 (ℏc)^3
where kB is Boltzmann’s constant, and T is the temperature. Plug in the values and show that the energy density is 4. 235 × 10 −^14 J/m^3. (b) The energy density of the CMB comes from the electromagnetic waves in the mi- crowave region of the electromagnetic spectrum. Recalling that the energy density of an electromagnetic wave is evenly split between the electric and magnetic fields (i.e., uE = uB ), determine the maximum values of the electric and magnetic fields of the CMB radiation.
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Solution
(a) Starting with the given expression, we just plug in the correct numbers
u =
π^2 k^4 B T 4 15 (ℏc)^3
π^2 (1. 38 × 10 −^23 )^4 (2.725)^4 15 (1. 05 × 10 − 34 × 2. 99 × 108 )^3
= 4. 235 × 10 −^14 J/m^3.
(b) Because u = 2uE = 0 E^2 , we just solve for the electric field. This gives, E =
u 0. Since E = cB, then B = (^1) c
u 0 =^
μ 0 u. So, using our results from part (b), we find E =
u 0
4. 235 × 10 −^14
8. 85 × 10 −^12
= 0. 0692 V /m,
and B =
μ 0 u =
4 π × 10 −^7 × 4. 235 × 10 −^14 = 2. 31 × 10 −^10 T. These are the maximum values.
- 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 accel- erates, and in accelerating it radiates energy. According to the Larmor formula, the power, P = dEdt , that is being radiated away as the electron accelerates is
dE dt
ke^2 c^3
a^2 ,
where a is the acceleration, and 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, using the Bohr model of the hydrogen atom.
(a) Recalling that, from the Bohr model of the hydrogen atom, the energy of the atom at radius r is given by E = −ke
2 2 r , take the time derivative of^ E^ and equate your expression to the Larmor formula above. Hint: from the chain rule, dEdt = dEdrdrdt. You should find that dr dt
r^2 a^2 c^3
(b) 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 (a) can be written as
dr dt
k^2 e^4 m^2 c^3
r^2
(c) 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 = 0.528 ˚A, what is the actual time in seconds?
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Solution
(a) Since E = −ke
2 2 r , then^
dE dt =^
dE dr
dr dt =^
ke^2 2 r^2
dr dt. Setting this equal to the Larmor for- mula gives ke 2 2 r^2
dr dt =^ −
2 3
ke^2 c^3 a
(^2). Canceling off the common factors gives the expected results. (b) The electrostatic force is just Coulomb’s law, F = ke
2 r^2 =^ ma, where^ m^ is the mass of the electron, and so a = ke
2 mr^2.^ Plugging in this result to part (a) gives dr dt =^ −
4 3
r^2 a^2 c^3 =^ −
4 3
r^2 c^3
ke^2 mr^2
= −^43 k
(^2) e 4 m^2 c^3
1 r^2 , as advertised. (c) Here we just separate and integrate, −^34 m (^2) c 3 k^2 e^4
r 0 r
(^2) dr = ∫^ t 0 dt, which gives
t =
m^2 c^3 r 03 4 k^2 e^4
Taking r 0 = 0.528 ˚A, and plugging in numbers gives t ≈ 1. 55 × 10 −^11 seconds! Clearly you can see why we need quantum mechanics!
