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 = 2 : L = 6 , Lz = 2 , ,0, - , -2 . (b) In each case cos
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 radii are roughly 3.6 fm, 5.3 fm, and 7.1 fm. (b) Usin
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 10-14 J (b) IDENTIFY and SET UP: The total energy of th
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 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk is constan
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 number of excess electrons needed to produce net charge
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 surface so
E = E cos dA = E cos A = (14 N/C )( cos 60 )
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 by Eq.(23.9). SET UP: Let the initial position of q2 b
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 charge -Q . Q PA and V = Ed . IDENTIFY and SET UP: C = 0 ,
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 objects in electrostatics, this is a huge amount of char
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 is two R/6 resistors in parallel. EXECUTE: The resista
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 + ( -3.85 104 m/s ) ^ ! ^ (a) B = (1.40 T ) i ! ! ! ^ ^ F
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 r ^ r ^ B= 0 = , since r = . 4 r 2 4 r 3 r ! ! 6 ^ and
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 coil. SET UP: The flux through a coil of N turns is = NBA
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 is the same as in part (a). Thus E1 = 0.270 V. dt EVALU
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(0.34 A) = 0.48 A. (c) Since the current is positive hal
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 (b) x = ct = (3.00 108 m/s)(8.61 yr)(3.156 107 s/yr) =
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 r = a = 39.4 . 11.5 cm EVALUATE: The angle of incidence
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. EXECUTE: The object and image are shown in Figure 34.1. s =
INTERFERENCE
35
35.1.
35.2.
IDENTIFY: Compare the path difference to the wavelength. SET UP: The separation between sources is 5.00 m, so for points between the sources the largest possible path difference is 5.00 m. EXECUTE: (a) For constructive interfer
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. Use this Eq.(36.2): sin = a equation to calculate . SET
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) = 1.36 103 m = 1.36 km. 37.3.
1 IDENTIFY and SET UP: The
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 when V0 = 0 is related to by = hf th .
EXECUTE: (a) From
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 = 0.155 nm (9.11 10-31 kg)(4.70 106 m/s)
m 9.11 10 -31 kg 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 = 1.2 10-67 J. 8(0.20 kg)(1.5 m) 2
1 2E 2(1.2 10-67 J) (b)