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
tx
2
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 electromagnetism in 1831. Besides the law named in
his h
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 for several years. He use two parallel metal electrod
es
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 wavelength of each photon is (using hc = 1240 eV nm )
=
hc
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. Adapting Eq. 42-21, there are
N Pu = M sam 1000 g NA = (6
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. (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 temperature T is given by
P( E ) = 1 e
( E E F )/ kT
+1
w
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 greatest value of ml is ml = 3 .
1 (b) Two states ( ms =
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 c. 10
34
J s 2.998 108 m / s
9
19
h 1240 eV nm. = J / e
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. Now E = hf = hc/, where h = 6.626 1034 Js is the Planc
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 at rest relative to the muon. Specifically, t0 = 2.200
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 angle is the same as for = b and m = 2. Thus, a = 2b = 2(35
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 indicates that 2 = 30 (which is what the problem gives as t
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 the mirror; the plane of its image is that same distanc
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 the eye is most sensitive is about 555 nm. (d) Using t
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 is +0.70 mWb as given in the problem statement. Since
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 the current (since UB i2), so this will occur after one
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 in the loop is clockwise. Thus, the current is to left
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 current must be from west to east to produce a field wh
Chapter 28
1. (a) The force on the electron is
r r rr FB = qv B = q vx ? + v y j Bx ? + By j = q ( vx By v y Bx ) k i i = 1.6 1019 C 2.0 106 m s ( 0.15 T ) 3.0 106 m s ( 0.030 T ) 14 = 6.2 10 N k.
r r Thus, the magnitude of FB is 6.2 1014 N, and FB points
Chapter 27
1. (a) The energy transferred is U = Pt =
2t
r+R
=
(2.0 V) 2 (2.0 min) (60 s / min) = 80 J . 1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s / min) = 67 J. U = i Rt = G JRt = G H+ R K H + 5.0 J 1.0