14
B05507025
2017.04.17
He-Ne
1. t x
1=E 1 cos ( t kx ) (1)
2
=2 v , k=
2=E 2 cos ( tkx + )( 2)
1 + 2 = E1 cos ( tkx )+ E 2 cos ( tkx + ) (3)
E1=E2=E
1 + 2 = 2 E cos cos ( t kx + ) ( 4)
2
cos
Chapter 41
1. (a) At absolute temperature T = 0, the probability is zero that any state with energy above the Fermi energy is occupied. (b) The probability that a state with energy E is occupied at te
Chapter 40
1. (a) Using Table 40-1, we find l = [ ml ]max = 4. (b) The smallest possible value of n is n = l max +1 l + 1 = 5.
1 (c) As usual, ms = 2 , so two possible values.
2. (a) For l = 3 , the g
Chapter 39
1. Since En L 2 in Eq. 39-4, we see that if L is doubled, then E1 becomes (2.6 eV)(2) 2 = 0.65 eV. 2. We first note that since h = 6.626 1034 Js and c = 2.998 108 m/s,
6 c.626 10 hc = 1602
Chapter 38
1. Let R be the rate of photon emission (number of photons emitted per unit time) of the Sun and let E be the energy of a single photon. Then the power output of the Sun is given by P = RE.
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
= 1
Ft I . Gt J HK
2 0
The proper time interval is measured by a clock
Chapter 36
1. (a) The condition for a minimum in a single-slit diffraction pattern is given by a sin = m, where a is the slit width, is the wavelength, and m is an integer. For = a and m = 1, the angl
Chapter 35
1. The index of refraction is found from Eq. 35-3: n= c 2.998 108 m s = = 156 . . . v 192 108 m s
2. Note that Snells Law (the law of refraction) leads to 1 = 2 when n1 = n2. The graph indi
Chapter 34
1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm. 2. The bird is a distance d2 in front of
Chapter 33
1. (a) From Fig. 33-2 we find the smaller wavelength in question to be about 515 nm. (b) Similarly, the larger wavelength is approximately 610 nm. (c) From Fig. 33-2 the wavelength at which
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 +(0.30 T)r2 where r = 0.020 m. The flux through the bottom
Chapter 31
1. (a) The period is T = 4(1.50 s) = 6.00 s. (b) The frequency is the reciprocal of the period:
f= 1 1 = = 167 105 Hz. . T 6.00s
(c) The magnetic energy does not depend on the direction of
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 f
Chapter 43
1. If MCr is the mass of a 52Cr nucleus and MMg is the mass of a 26Mg nucleus, then the disintegration energy is Q = (MCr 2MMg)c2 = [51.94051 u 2(25.98259 u)](931.5 MeV/u) = 23.0 MeV. 2. Ad
Chapter 44
1. The total rest energy of the electron-positron pair is
E = me c 2 + me c 2 = 2me c 2 = 2(0.511 MeV) = 1.022 MeV . With two gamma-ray photons produced in the annihilation process, the wav
14
B05507025
,
105.10.03()
(heat conduction)(thermal
conductivity)
1. A L
TATB A
B dQ/dt
T T A
dQ
=KA B
(1)
dt
L
K (1)
(1) L= x 0 T =T B T A 0
dQ
dT
=KA
(2)
dt
dx
(2)(1) K T x K
T K T
2.
Topic: Lenzs Law
b05507025 B05507055
1. Whos Lenz
Heinrich Friedrich Emil Lenz was a Russian physiciststudying chemistry and
physics at the University of Dorpat(In Estonia).
Lenz had begun studying e
Millikan Oil Drop Exp
eriment
Group 2
Whos Millikan
(Robert Millikan)
1910-1917
e
1916
1923
Background
J.J. Thomson has done the experiment about how to measure the char
ge of one electron fo
Chapter 2
1. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with x = v tc. (a)
Chapter 30
1. (a) The magnitude of the emf is
=
d B d = 6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV. dt dt
c
h
bg
(b) Appealing to Lenzs law (especially Fig. 30-5(a) we see that the current flow
Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B = 0i 2 r ,
i=
2 rB
0
=
2
b
4
m 39 10 6 T T m A
g c
h 16 A. =
(b) The
Chapter 14
1. The pressure increase is the applied force divided by the area: p = F/A = F/r2, where r is the radius of the piston. Thus p = (42 N)/(0.011 m)2 = 1.1 105 Pa. This is equivalent to 1.1 at
Chapter 13
1. The magnitude of the force of one particle on the other is given by F = Gm1m2/r2, where m1 and m2 are the masses, r is their separation, and G is the universal gravitational constant. We