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