Physics 2140: Hour Exam 3 Friday, July 6, 2007 Each Question is worth 10 points Circle 13 of the following 14 problems. I will only grade the circled problems! Please show your work in solving the the problems you choose. No credit will be awarded for answers without evidence of work supporting the answer. 1. The de Broglie wavelength of a 0.060 kg golf ball is 4.28 × 10−34 m. What is its speed? (h = 6.63 × 10−34 J⋅s) p = h / λ = mv ⇒ v = p / m = h / mλ = 25.8 m/s 2. X-rays of wavelength of 0.065 0 nm undergo Compton scattering from free electrons in carbon. What is the wavelength of photons scattered at 90.0° relative to the incident beam? (h = 6.63 × 10−34 J⋅s, me = 9.11 × 10−31 kg, c = 3.00 × 108 m/s, and 1 nm = 10−9 m) Δλ = λ − λ0 = h (1 − cos θ ) , λ0 = 6.5 ×10−11 m and cos 90D = 0 me c so, λ = 6.74 × 10−11 m = 0.0674 nm 3. In an x-ray diffraction experiment, using x-rays of wavelength λ = 0.500 × 10−10 m, a first-order maximum occurred at 5.00° off the crystal plane. Find the distance d between crystal planes. mλ = 2d sin θ , λ 2sin θ = d = 2.87 × 10−10 m 4. What is the highest frequency of the photons produced by a 90-kV x-ray machine? (h = 6.63 × 10−34 J⋅s) λmin = hc hc c , but c = f λ so, λmin = = eΔV eΔV f max Solving for f max gives: f max = eΔV = 2.2 ×1019 Hz h 5. A sodium vapor lamp has a power output of 300 W. If 590 nm is the average wavelength of the source, about how many photons are emitted per second? (h = 6.63 × 10−34 J⋅s, c = 3.00 × 108 m/s, and 1 nm = 10−9 m) E photon = hf = hc λ = 3.37 ×10−19 J/photon So, the total number of photons in 1 sec = 300/E photon = 8.90 ×1020 ≈ 1021 6. Blue light (λ = 460 nm) is incident on a piece of potassium (φ = 2.20 eV). What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10−34 J⋅s, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10−19 J, and 1 nm = 10−9 m) KE = hf − φ = hc λ − φ = 0.50 eV = 8.0 ×10-20 J 7. A satellite is powered by a small nuclear generator that puts out 15 W. How much matter is converted into energy over the 10 year life span of the generator? (1 year = 31577600 sec) 15W = 15 J/s, Energy = 15 W × 10y × 365.25 d/y × 24 h/d × 3600 s/h = mc2 = 4.73×109 J Then, solving for mass, m = Energy/c2 = 5.26×10-8 Kg = 53 μg 8. What is the total energy of a proton moving at a speed of 2.00 × 108 m/s? (proton mass is 1.67 × 10−27 kg and c = 3.00 × 108 m/s) E = γ mc 2 with γ = 1 1− β 2 = 1.34, so, E = 2.02 ×10−10 J = 1.26 GeV 9. An AC series circuit has 12.0 Ω resistance, 15.00 Ω inductive reactance and 10.00 Ω capacitive reactance. If an effective (rms) emf of 120 V is applied, what is the effective (rms) current value? Z = R 2 + ( X L − X C ) = 144 + 25 Ω = 13 Ω. Then, I = V / Z = 9.23 A 2 10. A series RLC AC circuit is at resonance. It contains a resistor of 30 Ω, a capacitor of 0.35 μF and an inductor of 90 mH. If an effective (rms) voltage of 150 V is applied, what is the effective (rms) current when the circuit is in resonance? At resonance Z=R, since XL=XC. Then, I=V/Z=V/R=5.0 A 11. A rocket ship is 80.0 m in length when measured before leaving the launching pad. What would its velocity be if a ground observer measured its length as 60.0 m while it is in flight? (c = 3.00 × 108 m/s) L = Lp / γ . So, γ = Lp / L = 80 / 60 = 4 / 3 = 1 1− β 2 Which yields, β = 0.66 = v / c, giving v = 1.98 ×108 m/s 12. The period of a pendulum is 2.0 s in a stationary inertial frame of reference. What is its period when measured by an observer moving at a speed of 0.60 c with respect to the inertial frame of reference? γ= 1 1− β 2 = 1.25 using v = 0.60c. τ p = 2.0 s Finally, τ = γτ p = 2.5 s 13. What value of inductance should be used in a series circuit with a capacitor of 1.8 nF when designed to radiate a wavelength of 35 m? (c = 3.00 × 108 m/s) 2π f 0 = 1 1 . Solving for inductance: L = . 2 LC ( 2π f ) C 2 ⎛ λ ⎞ 1 Using c = f λ we have: L = ⎜ = 1.9 × 10−7 H = 0.19 μ H ⎟ ⎝ 2π c ⎠ C 14. A radar pulse returns 3.0 × 10−4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object? (c = 3.00 × 108 m/s) Δx Or, Δx = cΔt = 9.0 × 104 m. But Δt is the time for the pulse to travel to the Δt object and return. So we need to divide Δx in half. Thus, the distance = 4.5 × 104 m = 45 km c= fλ = Physics 2140: Hour Exam 3 Answer Section Thursday, July 6, 2007 SHORT ANSWER 1. ANS: 26 m/s DIF: 2 2. ANS: 0.067 4 nm TOP: 27.6 The Dual Nature of Light and Matter DIF: 2 3. ANS: 2.87 × 10−10 m TOP: 27.5 The Compton Effect DIF: 2 4. ANS: 2.2 × 1019 Hz TOP: 27.4 Diffraction of X-Rays by Crystals DIF: 2 5. ANS: 1021 TOP: 27.3 X-Rays DIF: 2 6. ANS: 0.50 eV TOP: 27.2 The Photoelectric Effect and the Particle Theory of Light DIF: 2 7. ANS: 53 µg TOP: 27.2 The Photoelectric Effect and the Particle Theory of Light DIF: 2 8. ANS: 2.02 × 10−10 J TOP: 26.9 Relativistic Energy and the Equivalent of Mass and Energy DIF: 2 9. ANS: 9.23 A TOP: 26.9 Relativistic Energy and the Equivalent of Mass and Energy DIF: 2 10. ANS: 5.0 A TOP: 21.4 The RLC Series Circuit DIF: 2 11. ANS: 1.98 × 108 m/s TOP: 21.6 Resonance in a Series RLC Circuit DIF: 2 12. ANS: 2.5 s TOP: 26.6 Consequences of Special Relativity DIF: 2 13. ANS: 1.9 × 10−4 mH TOP: 26.6 Consequences of Special Relativity DIF: 2 14. ANS: 4.5 × 104 m TOP: 21.12 The Spectrum of Electromagnetic Waves DIF: 2 TOP: 21.12 The Spectrum of Electromagnetic Waves