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Chapter 32:

- 8: a) 100 mSv, b) 80 mSv, c) 15-30 mSv, depending on α energy
- 10: 2-4 Gy depending on the RBE of the α used.
- 16: Using RBE=20, 7.44 x 10
^{8}nuclei - 48: a) Z=90, N=142, b) product is
_{90}^{233}Th, c) two β decays produce_{91}^{233}Pa, and_{92}^{233}U successively, d) the U has N=141, e) half life is 1590 years, plenty long to have a bomb sit around for a while before you use it. - 49: Thermal output is 2570 MW, b) 8.04 x 10
^{19}fissions/sec, which is 2.53 x 10^{27}fissions/year, a total mass of U of 990 kg has fissioned - 52: 46 grams
- 54: 4.67 x 10
^{12}kWh, b) 4.67 x 10^{11}dollars

Chapter 31:

- 9: a) 4.6 fm, b) 0.61
- 10: 1 u gives 1.4924 x 10
^{-10}J, which is 931.5 MeV - 18: produces
_{20}^{40}Ca and an anti-neutrino (but I don't care much about neutrinos) - 20: produces
_{25}^{52}Mn - 22:
_{48}^{106}Cd - 24: This problem doesn't really ask a question, but the one I solved is... What is the daughter of the α decay of
^{226}Ra? The answer is..._{86}^{222}Rn - 34: a)
_{82}^{208}Pb, b) energy released = 33.05 MeV - 45: a) 0.50 mCi, b) this is 3 half lives, = 15.81 years
- 56: it decayed to 0.7 of its initial activity, which takes 2.71 years

Chapter 30:

- 7: Lyman series, n
_{f}=1, first line n_{i}=2, so λ = 121 nm - 9: yup, it works
- 11: since its energy is -0.85 eV, it will take 0.85 eV to ionize it
- 13: 2.12 x 10
^{-10}m

Chapter 29:

- 3: f=0.333 Hz so a) 2.2 x 10
^{-34}J, b) n = 2.26 x 10^{34}, c) tiny, tiny energy we cannot measure or see - 20: AM radio 1530 kHz a) 1.01 x 10
^{-27}J, b) 6.33 x 10^{-9}eV - 21: FM radio 90 MHz, a)5.97 x 10
^{-27}J, = 3.73 x 10^{-7}eV, b) at 50 kW of power this is 8.38 x 10^{29}photons/sec - 26: IR photon: a) E= 0.0827 eV, b) say 5 eV to break a molecule, this will take about 60 photons, c) γ-ray: 1.24 MeV, d) will break 250,000 molecules
- 35: at 650 kHz a 50kW radio station emits 1.16 x 10
^{32}photons/sec. Spread out over a sphere with one photon per cm^{2}per second, the sphere will have a radius of 3.04 x 10^{15}m, which is less than a light year, and stars that may have alien civilizations are much farther away than that. - 38: a) 1.67 x 10
^{-32}kg m/s, b) One reason is that you don't notice "impact" forces from them. - 39: a) 6.63 x 10
^{-23}kg m/s, b) 0.124 MeV, will kick butt on atomic electrons - 50: not relativity, too slow, so λ = 8.08 x 10
^{-11}m - 58: a) v = 728 m/s, b) use KE = q V, get V = 1.51 μV
- 86: F = 8.67 x 10
^{-6}N/m^{2}, so a = 8.67 x 10^{-5}m/s^{2}, so velocity gain each day is 7.5 m/s.

Chapter 28:

- 1: γ = 1.033 and 1.15
- 3: 5.96 x 10
^{-8}s - 6: v = 0.90 c
- 8: 0.24 c
- 13: 0.40 c
- 14: lives for 4.87 μs, and travels L
_{o}= 1390 m from earth observer's point of view. From its own perspective, it travels L= 433 m. - 20: 0.91 c
- 36: 1.35 x 10
^{-21}kg m/s - 40: 2.0 x 10
^{8}m/s - 44: 1.50 x 10
^{-10}J = 939 MeV - 45: 2.3 x 10
^{-30}kg - 47: a) mass lost = 1.11 x 10
^{27}kg, which is, b) 5.56 x 10^{-5}of the original mass - 69: Note: you need info from table 7.1, in chapter 7, about how much eneregy 1 kg of hydrogen will give you. Or, you might look up and find out that about 0.7% of the mass is lost, so 7 grams of mass turn into energy. Or you might find that each fusion of two protons releases 0.42 MeV of energy. Depending on which number you use, you might get somewhat different numbers from what I have here. The first two suggestions yield these answers:

a) 6.3 x 10^{11}kg/s of hydrogen is fusing. b) 1.4 x 10^{18}s, = 45 billion years, c) 4.4 x 10^{9}kg/s, d) 0.3 percent of its total mass

Chapter 27:

- 2: for visible range of 400nm - 700nm, λ in crown glass is 263nm - 460nm, for 380-760 nm it is 250-500 nm in crown glass
- 8: d = 1.22 μm
- 11: 577 nm
- 13: m=62 (set Sinθ=1, solve for m=62.5, but the biggest integer value of m=62)
- 23: 8990 lines/cm (solve for d, which is meters/line, convert to cm/line, and invert for lines/cm)
- 25: 5000 lines/cm gives d = 2.00 μm, λ = 707 nm
- 45: a) D=1.35 μm, b) θ=69.7
^{o} - 51: a) You can't use the small angle approximation, because the angles are 17.1306
^{o}and 17.1455^{o}, 0.0150^{o}different, b) And you cannot use the small angle approx to determine the separation, because it depends on the angles themselves. Find each y-value, then take the difference to get 0.273 mm - 58: 107 m
- 62: a) 2.237 x 10
^{-4}radians, which is 0.0128^{o}, b)5.8 km, c) 179 μm d) that is narrower than a pencil line, about the width of two hairs. - 86: 0.046 W/m
^{2} - 89: Final intensity is 1/4 the original intensity. If middle polarizer is removed, final intensity is zero.

Chapter 26:

- 2: 50.3 diopters
- 4: calculate the magnification, get 5.3 x 10
^{-6}, then calc the distance = 3.75 km - 6: 2.0 m
- 9: Original power = 51 diopters. so 51 = 1/x + 1/0.02, giving x = 1.0 m
- 16: -5 diopters is the prescription for contacts
- 17: -5.4 diopters is the prescription for glasses. (-5.6 D for glasses at 2 cm from eye)

Chapter 25:

- 2: Show that the change of direction equals 180 degrees
- 5: 2.25 x 10
^{8}m/s and 2.04 x 10^{8}m/s - 8: ice, n=1.31
- 13: n=1.457, so probably fused quartz (which is a fairly commonly used material
- 14: the time delay for two trips through the atmosphere is 5.86 x 10
^{-8}s, which is 2.29 x 10^{-6}% of the round trip time. - 22: 66.3
^{o} - 29: θ
_{red}= 46.5^{o}, θ_{violet}= 46.0^{o} - 36: 20.0 diopters
- 41: Screen is 3.43 m away. Mag = 33.3, so image is 80 cm x 120 cm.
- 42: a) -135 cm, that is, 135 cm from lens on the same side as the mole, b) m = 10, c) 50.0 mm image size
- 57: f = 36.0 cm
- 59: a) m = 0.111, b)image distance = -0.334 cm (behind the cornea), c) f = -0.376 cm, so R =2f = -0.752 cm

Chapter 24:

- 3: 1.50 x 10
^{5}V/m - 7: a) wavelengths from 188 to 556 m for AM, and 2.78 to 3.41 m for FM
- 12: λ = 0.25 μm, which is about the size of the smallest detail you could see with the UV microscope.
- 18: a) 9.46 x 10
^{15}m, b) 1.9 x 10^{22}m, c) 1.1 x 10^{26}m - 23: a) 193 nm has f = 1.55 x 10
^{15}Hz, the resolution ratio will be the ratio of the wavelengths, so for 400 nm light, 400/193 ≈2 two times better resolution. - 26: in 0.1 ns light goes 3 cm, which is the round trip uncertainty. Half of that is 1.5 cm, the uncertainty of the distance to the moon. b) 3.9 x 10
^{-9}percent - 32: Intensity, I = Power/area, 318 W/m
^{2}, b) B = 1.6 μT, c) E = 490 V/m - 42: a) 1.50 μm IR light, 200 W, in a 25.0 cm dia circle on your shoulder. Intensity, I=P/A = 4074 W/m
^{2}, b) E = 1750 V/m, c) B = 5.84 x 10^{-6}T, d) to heat 4.00 kg of shoulder requires 2.78 x 10^{4}J, and power = energy/time, so t = E/P = 139 seconds.

Chapter 23:

- 4: a) ccw, b) cw c) cw
- 9: a) 3.04 mV, b) pretty low power, not much temperature increase. Even if R = 0.001 mΩ the power is only 9 mW for a short time.
- 12: B = 0.425 T
- 17: 0.63 V, not much
- 27: a) I=0, b) I = cw, c) I = 0, d) I = ccw, e) I=0
- 50: a) 2.24, b) 0.447 (which is 1/2.24), c) loss is 20% of what it was before (0.447 squared)
- 67: L = 200 H, b) temperature increases by 5
^{o}C

Chapter 22:

- 2: a) right, b) Perp. out of page, c) down, to bottom of page, d) 0, e) left, f) up, to top of page
- 3: a) right, b) perp. into page, c) down, to bottom of page
- 10: 10.1
^{o}or 169.9^{o} - 14: 0.261 T, easily achieved even by permanent magnets.
- 32: a) left, b) perp. out of page, c) up to top of page
- 37: B = 1.80 T
- 48: τ = 0.471 N m, viewed from above, rotation will be cw.
- 57: F = 7.5 x 10
^{-5}N toward the bottom of the page, in line with the right side of the square. - 68: Need I
_{2}going down. I_{1}/I_{2}= 1/π - 84: need F
_{B}= mg, so you need to figure out m. I looked up the density of copper, which I won't tell you so you also have the pride of that accomplishment. Ultimately, I got I = 1730 A. If earth contributes a field of 3 x 10^{-5}T, this is 11% of the field contributed by the lower wire, so decrease its current by 11% to 1540 A.

Chapter 21:

- 3: Series gives the max R
_{s}=786 Ω, parallel gives the min,R_{p}=20.3 Ω - 6: a) I=0.400 A, P
_{24}=3.84 W, P_{96}=15.4 W, b) R_{p}=19.2 Ω, I=2.50 A, 2.0 A through R_{24}and 0.5 A through R_{96} - 16: 2.9994 V
- 23: a) 200 A, b) 10.0 V, c) 2000 W, d) 80 A, V
_{mot}= 4.0 V, P_{mot}= 320 W - 26: b) 1.67 A, c) 27.9 kW, d) 20 MΩ

For Resistors

(A) R

Note: R

(B) R

Note: R

(C) R

If instead they are capacitors

(A) C

(B) C

(C) C

Chapter 20:

- 4: 4.00 mA
- 9: 0.12 C, 7.5 x 10
^{17}electrons - 23: 0.20 mA
- 25: 0.322 Ω
- 27: the Al will be 1.24 times the diameter of the Cu wire
- 40: P = 2 x 10
^{12}W. But not for very long. - 49: beam power is 1500 W
- 56: $3.94 billion
- 69: a) = 2.08 x 10
^{5}A, for that current, b) the waste heat will be I^{2}R= 4.3 x 10^{10}W. c) but that is greater than the power output of the generator! So it cannot happen. d) Don't confuse the idea of how much power you want to get to the destination with how much power you need to produce at the source. - 73: 480 V
- 76: R = 576 Ω, power during melt down is 100 W.

Chapter 19:

- 2: eV=1/2 mv
^{2}, so v=1.2 x 10^{8}m/s - 5: 1940 V
- 7: a) 2 GJ, b) 765 kg of water can evaporate (mcΔT + mL
_{f}... remember?) - 13: yup, they are equivalent
- 17: a) E=2.5 MV, so no breakdown (no spark), b) at d=1.7 mm you get a big enough E-field.
- 29: Q = +0.833 μC
- 38: sketch so equipotentials are perp to E-field. Higher potential is in direction opposite the arrows of E-field
- 50: 20,000 V
- 61: C
_{net}= 11.4 μF. (a couple of intermediate combinations give 2.06 and 1.37 μF) - 66: series: 14.2 μC and 63.8 μJ, parallel: 84.6 μC and 381 μJ

Chapter 18:

- 1: a) 1.25 x 10
^{10}electrons b) 3.13 x 10^{12}electrons - 6: a) 1.88 x 10
^{6}electrons short of a full load, b) 1.88 x 10^{-10} - 16: 14.4 N away from Qa, (or in 1 sig fig, 10 N)
- 17: 3.45 x 10
^{16}m/s^{2} - 20: 1.04 nC
- 25: 85.9 cm right of negative charge, or, if both positive 10.9 cm left of +3 μC
- 30: 69.4 nC and 6.25 N/C
- 43: a) 4 x 10
^{7}N/C, b) 7 cm, in the middle, c)No, there is no suitable place 0-8 cm, d) The net charge in fig (a) is zero, but the net charge in fig (b) is +1, so E goes to zero faster in the case of fig. (a), and e) Add the electric fields due to the three charges for some point right of 11 cm. This is ugly algebra, which is why they suggest looking at the graph of the sum of the E-fields (if you have some way to make a graph of a function), then the graph of the sum of the E-fields crosses the zero at x=30.6 cm - 48: Vector addition! All the x-components cancel, and the net vector is 0.102 N in the negative y-direction
- 57: remember circular motion? F=ma=mv
^{2}/r, and v=rω, but F=kqQ/r^{2}, put it all together and get ω = 4.12 x 10^{16}rad/sec (and from v=rω get v=2.2 x 10^{6}m/s, if you are interested)

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