ATOMIC STRUCTURE
41
L = l (l + 1) . Lz = ml . l = 0, 1, 2,., n - 1. ml = 0, 1, 2,., l . cos = Lz / L .
41.1.
IDENTIFY and SET UP:
EXECUTE: (a) l = 0 : L = 0 , Lz = 0 . l = 1: L = 2 , Lz = ,0, - . l =
NUCLEAR PHYSICS
43
43.1.
(a) (b) (c)
28 14 85 37
Si has 14 protons and 14 neutrons. Rb has 37 protons and 48 neutrons. Tl has 81 protons and 124 neutrons.
205 81
43.2.
(a) Using R = (1.2 fm)A1 3 , the
PARTICLE PHYSICS AND COSMOLOGY
44
44.1.
(a) IDENTIFY and SET UP: Use Eq.(37.36) to calculate the kinetic energy K. 1 EXECUTE: K = mc 2 - 1 = 0.1547 mc 2 2 2 1- v / c
m = 9.109 10 -31 kg, so K = 1.27
SOUND AND HEARING
16
16.1.
IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) = v / f = (344
ELECTRIC CHARGE AND ELECTRIC FIELD
21
21.1.
(a) IDENTIFY and SET UP: Use the charge of one electron ( -1.602 10 -19 C) to find the number of electrons required to produce the net charge. EXECUTE: The
GAUSS'S LAW
22
^ E = E cos dA, where is the angle between the normal to the sheet n and the
22.1.
(a) IDENTIFY and SET UP:
electric field E . EXECUTE: In this problem E and cos are constant over the s
ELECTRIC POTENTIAL
23
ra = 0.150 m rb = (0.250 m) 2 + (0.250 m) 2 rb = 0.3536 m
23.1.
IDENTIFY: Apply Eq.(23.2) to calculate the work. The electric potential energy of a pair of point charges is given
CAPACITANCE AND DIELECTRICS
24
24.1.
24.2.
24.3.
Q Vab SET UP: 1 F = 10 -6 F EXECUTE: Q = CVab = (7.28 10 -6 F)(25.0 V) = 1.82 10 -4 C = 182 C EVALUATE: One plate has charge + Q and the other has char
CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE
25
25.1.
25.2.
IDENTIFY: I = Q / t . SET UP: 1.0 h = 3600 s EXECUTE: Q = It = (3.6 A)(3.0)(3600 s) = 3.89 104 C. EVALUATE: Compared to typical charges of o
DIRECT-CURRENT CIRCUITS
26
26.1.
26.2.
26.3.
IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP: Each of the three linear segments has resistance R/3. The circle
MAGNETIC FIELD AND MAGNETIC FORCES
27
27.1.
! IDENTIFY and SET UP: Apply Eq.(27.2) to calculate F . Use the cross products of unit vectors from Section 1.10. ! ^ j EXECUTE: v = ( +4.19 104 m/s ) i + (
SOURCES OF MAGNETIC FIELD
28
28.1.
! ^ EXECUTE: (a) r = ( 0.500 m ) i , r = 0.500 m ! ! ^ v r = vr^ i = -vrk j ^
! IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. ! ! ! ! ! qv r 0 qv
ELECTROMAGNETIC INDUCTION
29
29.1.
29.2.
IDENTIFY: Altering the orientation of a coil relative to a magnetic field changes the magnetic flux through the coil. This change then induces an emf in the co
INDUCTANCE
30
Apply Eq.(30.4). di (a) E2 = M 1 = (3.25 10-4 H)(830 A/s) = 0.270 V; yes, it is constant. dt
30.1.
IDENTIFY and SET UP: EXECUTE: (b) E1 = M
di2 ; M is a property of the pair of coils so
ALTERNATING CURRENT
31
31.1.
IDENTIFY: SET UP: EXECUTE:
i = I cos t and I rms = I/ 2.
The specified value is the root-mean-square current; I rms = 0.34 A.
(a) I rms = 0.34 A
31.2.
(b) I = 2 I rms = 2(
ELECTROMAGNETIC WAVES
32
32.1.
IDENTIFY: Since the speed is constant, distance x = ct. SET UP: The speed of light is c = 3.00 108 m/s . 1 yr = 3.156 107 s.
32.2.
x 3.84 108 m = = 1.28 s c 3.00 108 m/s
THE NATURE AND PROPAGATION OF LIGHT
33
33.1.
IDENTIFY: For reflection, r = a . SET UP: The desired path of the ray is sketched in Figure 33.1. 14.0 cm EXECUTE: tan = , so = 50.6 . r = 90 - = 39.4 and
GEOMETRIC OPTICS
34
y = 4.85 cm
Figure 34.1
34.1.
IDENTIFY and SET UP: Plane mirror: s = - s (Eq.34.1) and m = y / y = - s / s = +1 (Eq.34.2). We are given s and y and are asked to find s and y. EXECU
DIFFRACTION
36
36.1.
IDENTIFY: Use y = x tan to calculate the angular position of the first minimum. The minima are located by m , m = 1, 2,. First minimum means m = 1 and sin 1 = / a and = a sin 1. U
RELATIVITY
37
Figure 37.1
37.1.
IDENTIFY and SET UP: Consider the distance A to O and B to O as observed by an observer on the ground (Figure 37.1).
(b) d = vt = (0.900) (3.00 108 m s) (5.05 10-6 s) =
PHOTONS, ELECTRONS, AND ATOMS
38
h f - . The e e
38.1.
IDENTIFY and SET UP: The stopping potential V0 is related to the frequency of the light by V0 = slope of V0 versus f is h/e. The value fth of f w
THE WAVE NATURE OF PARTICLES
39
hc
39.1.
IDENTIFY and SET UP: EXECUTE: (a) =
=
h h = . For an electron, m = 9.11 10 -31 kg . For a proton, m = 1.67 10 -27 kg . p mv
6.63 10-34 J s = 1.55 10-10 m = 0.1
QUANTUM MECHANICS
40
n2h 2 . 8mL2
40.1.
IDENTIFY and SET UP: The energy levels for a particle in a box are given by En = EXECUTE: (a) The lowest level is for n = 1, and E1 =
(1)(6.626 10-34 J s) 2 = 1