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
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
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)
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
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
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
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
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 =
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
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) =
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
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
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
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
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
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
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
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 ,
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
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 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
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