Chapter 42
1. Kinetic energy (we use the classical formula since v is much less than c) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for lithium and Z = 90 for thorium; the charges on those nuclei are therefore 3
Chapter 41
1. According to Eq. 41-9, the Fermi energy is given by
EF
F3 I h n G 2 J m HK 16
2/3 2
2/3
where n is the number of conduction electrons per unit volume, m is the mass of an electron, and h is the Planck constant. This can be written EF = An2/
Chapter 40
1. The magnitude L of the orbital angular momentum L is given by Eq. 40-2: L ( 1) . On the other hand, the components Lz are Lz m , where m ,. . Thus, the semi-classical angle is cos Lz / L . The angle is the smallest when m , or
cos . cos 1 (
Chapter 39
1. According to Eq. 39-4, En L 2. As a consequence, the new energy level E'n satisfies
En L En L
L F I F I 1, GJ G J 2 HK HK L
2 2
which gives L 2 L. Thus, the ratio is L / L 2 1.41. 2. (a) The ground-state energy is
2 6.63 1034 J s h2 2 E1 n 3
Chapter 38
1. (a) With E = hc/min = 1240 eVnm/min = 0.6 eV, we obtain = 2.1 103 nm = 2.1 m. (b) It is in the infrared region. 2. Let and solve for v:
v
1 hc me v 2 E photon 2
2 hc 2 hc 2 2 hc c c 2 me me c me c 2
5 3
2.998 108 m/s
588 nm 511 10 eV 8.6
Chapter 37
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
1 / 1 2 , and = v/c), we obtain
Ft I 1 G J. Ht K
2 0
The proper time interval is measured by a clock at rest relative to the muon. Specifi cally, t0 = 2.2000 s.
Chapter 36
1. (a) We use Eq. 36-3 to calculate the separation between the first (m1 = 1) and fifth (m2 5) minima:
D m D y D sin D m m2 m1 . a a a
Solving for the slit width, we obtain
400 mm 550 106 mm 5 1 D m2 m1 a 2.5 mm . y 0.35 mm
(b) For m = 1,
6 m 1
Chapter 35
1. The fact that wave W2 reflects two additional times has no substantive effect on the calculations, since two reflections amount to a 2(/2) = phase difference, which is effectively not a phase difference at all. The substantive differen ce be
Chapter 34
1. The bird is a distance d2 in front of the mirror; the plane of its image is that same distance d2 behind the mirror. The lateral distance between you and the bird is d3 = 5.00 m. We denote the distance from the camera to the mirror as d1, an
Chapter 33
1. Since , we find f is equal to
8 9 c c (3.0 10 m/s)(0.0100 10 m) 2 7.49 109 Hz. (632.8 109 m) 2
2. (a) The frequency of the radiation is
f c 3.0 108 m / s 4.7 103 Hz. (1.0 105 )(6.4 106 m)
(b) The period of the radiation is
T
1 1 212 s 3 mi
Chapter 32
1. We use
6 n 1
Bn 0 to obtain
5
B 6 Bn 1Wb 2 Wb 3 Wb 4 Wb 5 Wb 3 Wb .
n 1
2. (a) The flux through the top is +(0.50 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must b
Chapter 31
1. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is
U
Q 1.17 106 J. 6 2C 2 3.60 10 F
2
2.90 106 C
2
(b
Chapter 30
1. The flux B BA cos does not change as the loop is rotated. Faradays law only leads to a nonzero induced emf when the flux is changing, so the result in this instance is zero. 2. Using Faradays law, the induced emf is
d r2 d BA dB dA dr B B 2
Chapter 29
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by i B 0 . 2 r With r = 20 ft = 6.10 m, we have
4 10 B
7
T m A 100 A
2 6.10 m
3.3 10
6
T 3.3 T.
(b) This is about one-si
Chapter 28
1. (a) Equation 28-3 leads to
v
eB sin
FB
1.60 10 C2.60 10 Tsin 23.0
19 3
6.50 1017 N
4.00 105 m s .
(b) The kinetic energy of the proton is
K
2 121 mv 1.67 10 27 kg 4.00 10 5 m s 1.34 10 16 J , 2 2
which is equivalent to K = (1.34 10 16 J) /
Chapter 27
1. (a) Let i be the current in the circuit and take it to be positive if it is to the left in R1. We use Kirchhoffs loop rule: 1 iR2 iR1 2 = 0. We solve for i:
i R1 R2
1 2
12 V 6.0 V 0.50 A. 4.0 8.0
A positive value is obtained, so the current
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is t
Chapter 25
1. (a) The capacitance of the system is
C
q 70 pC 3.5 pF . V 20 V
(b) The capacitance is independent of q; it is still 3.5 pF. (c) The potential difference becomes
V
q 200 pC 57 V . C 3.5 pF
2. Charge flows until the potential difference acros
Chapter 24
1. (a) An ampere is a coulomb per second, so
s Ch 5 84 A h 84 3600 3.0 10 C. s h
(b) The change in potential energy is U = qV = (3.0 105 C)(12 V) = 3.6 106 J. 2. The magnitude is U = eV = 1.2 109 eV = 1.2 GeV. 3. If the electric potential is ze
Chapter 23
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
2 E A EA cos 1800 N C 3.2 103 m cos145 1.5 102 N m 2 C.
2. We use E dA and note th
Chapter 22
1. We note that the symbol q2 is used in the problem statement to mean the absolute value of the negative charge that resides on the larger shell. The following sketch is for q1 q2 .
The following two sketches are for the cases q1 > q2 (left fi
Chapter 21
1. The magnitude of the force of either of the charges on the other is given by
F 1 q Qq 4 0 r2
bg
where r is the distance between the charges. We want the value of q that maximizes the function f(q) = q(Q q). Setting the derivative dF / dq equ
Chapter 20
1. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion is
W p dV n RT
V1 V2 V2
V1
dV V n RT ln 2 . V
Chapter 19
1. Each atom has a mass of m = M/NA, where M is the molar mass and NA is the Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9 103 kg/mol. Therefore, 7.50 1024 arsenic atoms have a total mass of (7.50 1024) (74.9 103 kg/mol)/(6
Chapter 18
1. From Eq. 18-6, we see that the limiting value of the pressure ratio is the same as the absolute temperature ratio: (373.15 K)/(273.16 K) = 1.366. 2. We take p3 to be 80 kPa for both thermometers. According to Fig. 18-6, the nitrogen thermome
Chapter 17
1. (a) The time for the sound to travel from the kicker to a spectator is given by d/v, where d is the distance and v is the speed of sound. The time for light to travel the same distance is given by d/c, where c is the speed of light. The dela
Chapter 16
1. Let y1 = 2.0 mm (corresponding to time t1) and y2 = 2.0 mm (corresponding to time t2). Then we find and kx + 600t1 + = sin1(2.0/5.0) kx + 600t2 + = sin1(2.0/5.0) . Subtracting equations gives 600(t1 t2) = sin1(2.0/5.0) sin1(2.0/5.0). Thus we
Chapter 15
1. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a turning point (that is, when x = +xm or x = xm). Consider that it starts at x = +xm and we are told that t = 0.25 second elapses until the object reac
Chapter 14
1. Let the volume of the expanded air sacs be Va and that of the fish with its air sacs collapsed be V. Then m m fish fish 1.08 g/cm3 and w fish 1.00 g/cm3 V V Va where w is the density of the water. This implies
fishV = w(V + Va) or (V + Va)/V