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Unformatted text preview: 25.38 The source is approaching you, so the situation is actually a “blue shift.” In your haste, you have
seen the trafﬁc light shift from a frequency of e 3.00x108 m/s . l4
=——=——~——._:4. 4 1
V0 A0 620x10“9m 8 X 0 HZ to a
p=§2W=356x1014Ha 548 X 10‘9 In 'U
1+; 9 9o 1; i22wa
we 6:» wee—em. 2e
\1—«— C V‘+VD
C 2% v : (5.56 x 1014 Hz)? a (4.84 x 1014 Hz)2 7
(5.56 x1014 Hz)2 + (4.64 x 1014 Hz)?” 014C 2 4'2 X 10 W3 In kin/i1 this is _ 7 ' km 3600s
771—.(42x10 iii/2:)(103m T al.5x108km/h. CHAPTER 25. THE SPECIAL THEORY OF RELATIVITY The problem is. given to us in terms orr wavelengths rather than frequencies, so We ’11 begin by rewriting
this relation in terms of wavelengths A and A0. Since G 2 VA, We have 1/ 2 5, and also 120 2 gm. Hence7 A0
02 2 .1—‘—'SO C c
A A0 02! 2 A0 '1}
(1) Te 1—“ 62 liVe’li ﬁrst solve this equation for e, and
656.382 nm, and c = 3.00 x 105 m/s. Begin by Squaring both sides of (1), and then turn the crank. _/\0 2 ’U‘O' U2 A0 2 ‘
(Y) “hence“(7 Tr then substitute the known values of A0 2 556.282 nm, A : 1M 656.282nm 2
65638211111 .. 25.46
a) The rest energy is Brest : me2 = {5.0 ><10#3 kg)(3.00 x 208 m/s)2 2 4.5 X 1014 J. 981 b} The relativistic kinetic energy is given by Equation 25.?5 on page 1184 of the text. KE=(’{—l)mc2 ———1—v——2m_l e202,
‘"  “(3) where m = 1.0 kg is new the mass. Set this equal to H the rest energy of the penny and then solve fer v. 1 4.5x1014J= W4 meg: , 1 2_1 (1.0kg){3.00x108m/s)2
1—(3) 1(5) 
=> 1 24230050
ME) '
was, i 221.0050
ME). :3 v : 0.0995c = 2.99 x 107 m/S. 25.49 First ﬁnd the relativistic factor l l
7 v2 x/l m (0.600)?
_.C? The kinetic energy is 3:3 2 (7 ~_ mm2 = (1.25 s 1){9.11 x 1031 kg)(3.00 x 108 m/af’ = 2.05 x 10—14 J. The total relativistic energy is E = 7ch = 1.25911 x 1031 kg)(3.00 x108 III/S)? = 1102 x 10"13 J. The magnitude of the momentum is p : 77m; .4 12509.11 x 10”31kg)0.600(3_00 x105 m/s) = 2.05 x io~22 kgm/s. As a check, evaluate both sides of the equation The ieft hand side is me : (1.02 x 1013 J}2 = 1.04 x 2025 J2.
The right hand side is RHS : (2.05 >910“22 kgm/S)2(3.OO x 108 111/3)? + (9.11 x 1031 kg)2(3.00 x 105 m/s) Close enough! 4 = 1.05 x 1926 32. 25.53 a) The condition is E = MOErest rs» vmcz = 1.10mi:2 => 7:1.10 => W 2 1.10.
1“ "a"
c
Square both sides of the last equation and solve for 1:.
1 2 '02?— l :21]: 1— 1 c:0.417c:1.25x108m/s. u? E (1'10) :> 1— (2—2 “ (map (1.10)2 1* 25 210‘} CHAPTER 25. THE SPECIAL THEORY OF RELATIVITY b) By the OWE theorem, the work done is equal to the change in the kinetic energy: W:AKE=KEf—KE;=(7—l)mc2—GJ 2 (1.10~1)(1.67 x 1027 kg)(3.80 x 10*3 111/3)? 2 1.5 x 1011J. Note that when expressed in units of electromvolts, V .
W x (1.5 X 10_11 = 9.4 X 107 8V. So, to reach this kinetic energy, the proton would have to be accel erated through a potential difference of
94 million volts. 25.60 a) From the CW’E theorem, the work done is the change in kinetic energy. Since the electron is starting
at rest, the change in kinetic energy is its total kinetic energy KEOIG when moving at v : 0.5006. According to Equation 25.15 on page 1184 of the text, 1
B3530.6 = (T — Umcz : —‘—M— f 1 mi:2 1— (3)2 1
a (——_ — 1) (9.11 ><10"31 kg)(3.00 x 108 m/S)2 = 2.05 x 1914 J. \/i w 0.6002 Converted to keV this is Keg... = 2.05 x 10"14 J :(2115 x 10*” J) (10128;) = :28 ken b) The additionai work is‘the change in kinetic energy when going from a speed of 0.60% to a speed of
0.800c. We’ve aiready computed Kline, so now compute 1
.I{E0_g:(7—i)mc2: m~l me? 1(i)2 : (v1 — 0.8002 e 1) (9.11 x10‘31kg)(3.00 x 108 m/S)2 = 5.47 x 10"14 J. Converted to keV this is V keV
: .4 1 14J : .4 1 "14 ————em.._ : .
REM 5 7X {1 (5 7x G J) (1602 X1049 J 103 eV 341keV The additional work, in joules, is are: : K1303 u Keg6 m 5.47 X 1014 J M 2.05 x 10“14 J = 3.42 x 10~14 J. Measured in keV, this is axe : Keg8 H new = 341 kev * 128 keV 2 213 keV. c) The ratio of kinetic energies is ~14
KEos H 5.47 x 10 J x 2.67. K1303 _ 2.05 x 1031 J . So the electron has 2.67 times as much kinetic energy at a speed of 0.8000 as it doesat 0.600c. As the
electron goes faster, it takes more and more energy to get the same increase in Speed: Eventually, as 1) —> c, it takes an inﬁnite amount of energy. 26.2 Use W'ien’s displacement law, Equation 26.17 on page 1210 of the text: 2 ' . .2 “*2 . AmaXT 2 0.28978 X 10‘2 mK :4 Am : 1% Your normal body temperature is
i
i 0.28978 x 107.2 mK _ 0.28978 x 10*"2 mK _ u 1 we 
AniaHT W—QﬁSXlO In This is in the infrared region of the electromagnetic _speetrum. Many interesting applications are based on the fact that living organisms
of infrared radiation. For example Earth’s surface; night vision goggles use infrared radiation to enable people to “
(is, in the absence of Visible licrht); etc. 26.8 Let N be the rate at which photons are emitted (photons per unit time), E be the energy per photon,
and P be the power output. Then, since power is energy per unit time, P 2 N E .
The energy of each photon is by, so E = by. Thus P 50x103W ‘85X1029 h
“*m _' poton/s.
13:th z; N—hv —WW ...
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This note was uploaded on 10/01/2009 for the course PHYS 102 taught by Professor Henry during the Spring '09 term at Johns Hopkins.
 Spring '09
 HENRY
 Physics

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