Type:
Instructor Manual
Resource:
Modern Physics
Edition:
4th Edition
Author(s):
Kenneth S. Krane
Chapter 1 1. (a) Conservation of momentum gives px ,initial = px ,final , or mH vH,initial + mHe vHe,initial = mH vH,final + mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
vHe,final =
mH (vH,initial − vH,final )
mHe (1.674 ×10−27 kg)[1.1250 ×107 m/s − (−6.724 ×106 m/s)] = = 4.527 ×106 m/s −27 6.646 ×10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic collision. Conservation of energy then gives K initial = K final , or 1 2
2 2 2 2 mH vH,initial + 12 mHe vHe,initial =12 mH vH,final + 12 mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
vHe,final = =
2 2 − vH,final mH (vH,initial )
mHe (1.674 ×10−27 kg)[(1.1250 ×107 m/s) 2 − (−6.724 ×106 m/s) 2 ] = 4.527 ×106 m/s −27 6.646 ×10 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum gives: : = px ,initial px ,final = mHe vHe,initial mHe vHe,final cos θ He + mO vO,final cos θ O p y ,initial p= 0 mHe vHe,final sin θ He + mO vO,final sin θ O = y ,final : From the second equation,
sin θ He m v (6.6465 ×10−27 kg)(6.636 ×106 m/s)(sin 84.7°) vO,final = 2.551×106 m/s − He He,final = − = (2.6560 ×10−26 kg)[sin(−40.4°)] mO sin θ O (b) From the first momentum equation,
Chapter 1
Page 1
vHe,initial =
mHe vHe,final cos θ He + mO vO,final cos θ O
mHe (6.6465 × 10 kg)(6.636 × 106 m/s)(cos 84.7°) + (2.6560 × 10−26 kg)(2.551 × 106 m/s)[cos(−40.4°)] = 6.6465 × 10−27 kg = 8.376 × 106 m/s −27
3. (a) Using conservation of momentum for this one-dimensional situation, we have px ,initial = px ,final , or
mHe vHe + mN vN = mD vD + mO vO Solving for vO with vN = 0 , we obtain
mHe vHe − mD vD (3.016 u)(6.346 ×106 m/s) − (2.014 u)(1.531×107 m/s) vO = = = −7.79 ×105 m/s mO 15.003 u (b) The kinetic energies are: 2 1.008 ×10−13 J K initial =12 mHe vHe + 12 mN vN2 =12 (3.016 u)(1.6605 ×10−27 kg/u)(6.346 ×106 m/s) 2 = K final = 12 mD vD2 + 12 mO vO2 = 12 (2.014 u)(1.6605 ×10−27 kg/u)(1.531×107 m/s) 2 3.995 ×10−13 J + 12 (15.003 u)(1.6605 ×10−27 kg/u)(7.79 ×105 m/s) 2 =
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic energy. The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds v1 and v2 . Conservation of momentum along the x direction ( p x ,initial = p x ,final ) gives
0= m1v1 − m2 v2 or
v1 = v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 + K 2 = 92.2 keV Because v1 = v2 , it follows that K= K= 46.1 keV , so 1 2 = v
2 K1 = m1
2(46.1 ×103 eV)(1.602 ×10−19 J/eV) = 1.49 ×106 m/s (4.00 u)(1.6605 ×10−27 kg/u)
v2= v= 1.49 ×106 m/s 1
Chapter 1
Page 2
5. (a) The kinetic energy of the electrons is −31 2 1 1 Ki = kg)(1.76 ×106 m/s) = 14.11×10−19 J 2 mvi = 2 (9.11× 10
In passing through a potential difference of ∆V = Vf − Vi = +4.15 volts , the potential energy of the electrons changes by
∆U =∆ q V =− ( 1.602 ×10−19 C)( + 4.15 V) = −6.65 ×10−19 J Conservation of energy gives K i + U i = K f + U f , so
K= K i + (U i − U f = ) K i − ∆U= 14.11×10−19 J + 6.65 ×10−19 = J 20.76 ×10−19 J f 2Kf = m
= vf
2(20.76 ×10−19 J) = 2.13 ×106 m/s −31 9.11× 10 kg
(b) In this case ∆V = −4.15 volts, so ∆U = +6.65 × 10−19 J and thus
K= K i − ∆U= 14.11×10−19 J − 6.65 ×10−19 = J 7.46 ×10−19 J f = vf
2Kf = m
2(7.46 ×10−19 J) = 1.28 ×106 m/s −31 9.11×10 kg
6. (a) ∆x A =v∆t A =(0.624)(2.997 ×108 m/s)(124 ×10−9 s) =23.2 m (b) ∆xB =v∆t B =(0.624)(2.997 ×108 m/s)(159 ×10−9 s) =29.7 m 7. With T = 35°C = 308 K and= P 1.22 atm = 1.23 ×105 Pa ,
N P 1.23 × 105 Pa = = = 2.89 × 1025 atoms/m3 -23 V kT (1.38 × 10 J/K)(308 K) so the volume available to each atom is (2.89 × 1025/m3)−1 = 3.46 × 10−26 m3. For a spherical atom, the volume would be 4 3
−10 4 π R3 = m)3 =× 1.50 10−30 m3 3 π (0.710 × 10
The fraction is then
1.50 ×10−30 = 4.34 ×10−5 −26 3.46 ×10 Chapter 1
Page 3
8. Differentiating N(E) from Equation 1.23, we obtain
dN 2 N 1 1 −1/ 2 − E / kT 1 − E / kT = E e + E1/ 2 − e 3/ 2 2 dE π (kT ) kT To find the maximum, we set this function equal to zero: 1 1 E E −1/ 2 e − E / kT − 0 = 3/ 2 π (kT ) 2 kT
2N
Solving, we find the maximum occurs at E = 12 kT . Note that E = 0 and E = ∞ also satisfy the equation, but these solutions give minima rather than maxima. 9. For this case kT= (280 K)(8.617 ×10−5 eV/K) = 0.0241 eV . We take dE as the width of the interval (0.0012 eV) and E as its midpoint (0.0306 eV). Then dN = N ( E ) dE =
10.
2N
1 (0.0306 eV)1/2 e − (0.0306 eV)/(0.0241 eV) (0.0012 eV) = 1.8 ×10−2 N 3/2 π (0.0241 eV)
(a) From Eq. 1.33, ∆Eint=
5 2
nR ∆T=
5 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 3.21×103 J
7 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 4.50 ×103 J
(b) From Eq. 1.34, ∆Eint=
7 2
nR ∆T=
(c) For both cases, the change in the translational part of the kinetic energy is given by Eq. 1.31: ∆Eint=
11.
3 2
nR ∆T=
5 2
(2.37 moles)(8.315 J/mol ⋅ K)(65.2 K)= 1.93 ×103 J
After the collision, m 1 moves with speed v1′ (in the y direction) and m 2 with speed v2′ (at an angle θ with the x axis). Conservation of energy then gives E initial = E final : 1 2
1 ′2 1 ′2 m1v12 = 2 m1v1 + 2 m2 v2
or
v2 = v1′2 + 3v2′2
Conservation of momentum gives:
Chapter 1
Page 4
= p x ,initial p= m1v1 m2v2′ cos θ = or v 3v2′ cos θ x ,final : p y ,initial = p y ,final : 0 = m1v1′ − m2v2′ sin θ or
v1′ = 3v2′ sin θ
We first solve for the speeds by eliminating θ from these equations. Squaring the two momentum equations and adding them, we obtain v 2 + v1′2 = 9v2′2 , and combining this result with the energy equation allows us to solve for the speeds: = v1′ v= / 2 and v2′ v / 6 By substituting this value of v2′ into the first momentum equation, we obtain cos = θ
12.
2/3
= θ 35.3°
or
The combined particle, with mass m′ = m1 + m2 = 3m , moves with speed v′ at an angle θ with respect to the x axis. Conservation of momentum then gives: p x ,initial p= m1v1 m′v′ cos θ = or v 3v′ cos θ = x ,final : 4 p y ,initial p= m2v2 m′v′ sin θ = or 3v′ sin θ = y ,final : 3v
We can first solve for θ by dividing these two equations to eliminate the unknown v′:
tan= θ
4 3
or
= θ 53.1°
Now we can substitute this result into either of the momentum equations to find v′ = 5v / 9
The kinetic energy lost is the difference between the initial and final kinetic energies: 26 1 K initial − K final = 12 m1v12 + 12 m2v22 − 12 m′v′2 = 12 mv 2 + 12 (2m )( 23 v ) 2 − 12 (3m )( 95 v ) 2 = 27 ( 2 mv 2 ) 2 17 1 The total initial kinetic energy is 12 mv 2 + 12 (2m)( 32 v) 2 = . The loss in kinetic 9 ( 2 mv )
energy is then 26 51 = 51% of the initial kinetic energy. 13.
(a) Let v1 represent the helium atom that moves in the +x direction, and let v2 represent the other helium atom (which might move either in the positive or negative x direction). Then conservation of momentum ( p x ,initial = p x ,final ) gives
mv = m1v1 + m2 v2 or Chapter 1
2v = v1 + v2 Page 5
where v2 may be positive or negative. The initial velocity v is v =
2K = m
2(40.0 × 103 eV)(1.602 × 10-19 J/eV) = 9.822 × 105 m/s -27 (8.00 u)(1.6605 × 10 kg/u)
The energy available to the two helium atoms after the decay is the initial kinetic energy of the beryllium atom plus the energy released in its decay: K + 92.2 keV = 12 m1v12 + 12 m2 v22 = 12 m1v12 + 12 m2 (2v − v1 ) 2
where the last substitution is made from the momentum equation. Solving this quadratic equation for v1 , we obtain v1 =2.47 ×106 m/s or − 0.508 ×106 m/s . Because we identified m 1 as the helium moving in the positive x direction, it is identified with the positive root and thus (because the two heliums are interchangeable in the equation) the second value represents the velocity of m 2 : v1 =× 2.47 106 m/s, v2 = −0.508 ×106 m/s
(b) Suppose we were to travel in the positive x direction at a speed of v = 9.822× 105 m/s, which is the original speed of the beryllium from part (a). If we travel at the same speed as the beryllium, it appears to be at rest, so its initial momentum is zero in this frame of reference. The two heliums then travel with equal speeds in opposite directions along the x axis. Because they share the available energy equally, each helium has a kinetic energy of 46.1 keV and a speed of 2 K = / m 1.49 × 106 m/s , as we found in Problem 4. Let’s represent these velocities in this frame of reference as v1′ = +1.49 ×106 m/s and v2′ = −1.49 ×106 m/s . Transforming back to the original frame, we find v1 = v1′ + v = 1.49 ×106 m/s + 9.822 ×105 m/s = 2.47 ×106 m/s v2 =v2′ + v =−1.49 ×106 m/s + 9.822 ×105 m/s =−0.508 ×106 m/s
14.
(a) Let the second helium move in a direction at an angle θ with the x axis. (We’ll assume that the 30° angle for m 1 is measured above the x axis, while the angle θ for m 2 is measured below the x axis. Then conservation of momentum gives:
: mv m1v1 cos 30° + m2 v2 cos θ = px ,initial px ,final =
or
3 2v= − v1 v2 cos θ 2
1 or = v1 v2 sin θ 2 We can eliminate the angle θ by squaring and adding the two momentum equations: = p y ,initial p y= m1v1 sin 30° − m2 v2 sin θ ,final : 0
Chapter 1
Page 6
4v 2 + v12 − 2 3vv1 = v22 The kinetic energy given to the two heliums is equal to the original kinetic energy 2 1 2 mv of the beryllium plus the energy released in the decay: 1 2
mv 2 + 92.2 keV = 1 2
1 2
2 m1v12 + 12 m2 v= 2
1 2
m1v12 + 12 m2 (4v 2 + v12 − 2 3vv1 )
(m1 + m2 )v12 − 3m2 vv1 + (2m2 v 2 − 12 mv 2 − 92.2 keV) = 0
Solving this quadratic equation gives v1 = 2.405 ×106 m/s, − 0.321×106 m/s
Based on the directions assumed in writing the momentum equations, only the positive root is meaningful. We can substitute this value for v1 into either the momentum or the energy equations to find v2 and so our solution is: v1 = 2.41×106 m/s, v2 = 1.25 ×106 m/s
The angle θ can be found by substituting these values into either of the momentum equations, for example v1 2.41×106 m/s θ sin −1 = sin −1 = = 74.9° 2v2 2(1.25 ×106 m/s) (b) The original speed of the beryllium atom = is v 2 K= / m 1.203 × 106 m/s . If we were to view the experiment from a frame of reference moving at this velocity, the original beryllium atom would appear to be at rest. In this frame of reference, in which the initial momentum is zero, the two helium atoms are emitted in opposite directions with equal speeds. Each helium has a kinetic energy of 46.1 keV and a ′ v= ′ 1.49 ×106 m/s . Let φ represent the angle that each of the helium speed of v= 1 2 atoms makes with the x axis in this frame of reference. Then the relationship between the x components of the velocity of m 1 in this frame of reference and the original frame of reference is
v1 cos = 30° v1′ cos φ + v and similarly for the y components
v1 sin 30° =v1′ sin φ We can divide these two equations to get Chapter 1
Page 7
v′ cos φ + v cot 30° = 1 v1′ sin φ
which can be solved to give φ = 53.8°. Using this value of φ, we can then find v1 2.41×106 m/s . We can also write the velocity addition equations for m 2 : =
v2 cos θ = −v2′ cos φ + v
and
v2 sin θ = −v2′ sin φ
which describe respectively the x and y components. Solving as we did for m 1 , we find v2 = 1.25 ×106 m/s and θ = 74.9° .
15.
(a) With K = 23 kT , ∆K =
3 2
k ∆T =
3 2
(1.38 × 10−23 J/K)(80 K) = 1.66 × 10−21 J = 0.0104 eV
(b) With U = mgh ,
= h
16.
U 1.66 × 10−21 J = = 2550 m mg (40.0 u)(1.66 × 10−27 kg/u)(9.80 m/s2 )
We take dE to be the width of this small interval: dE = 0.04kT − 0.02kT = 0.02kT , and we evaluate the distribution function at an energy equal to the midpoint of the interval (E = 0.03kT):
dN N ( E ) dE = = N N
17.
2
1 (0.03kT )1/ 2 e − (0.03kT ) / kT (0.02kT ) 3.79 ×10−3 = 3/ 2 π (kT )
If we represent the molecule as two atoms considered as point masses m separated by a distance 2R, the rotational inertia about one of the axes is I x′ = mR 2 + mR 2 = 2mR 2 . On average, the rotational kinetic energy about any one axis is 12 kT , so 1 2
= ω x′
I x′ω x2′ = 12 kT and
kT = I x′
Chapter 1
kT = 2mR 2
(1.38 ×10−23 J/K)(300 K) = 4.61×1012 rad/s −27 −9 2 2(15.995 u)(1.6605 ×10 kg/u)(0.0605 ×10 m)
Page 8
Chapter 2 1. Your air speed in still air is (750 km)/(3.14 h) = 238.8 km/h. With the nose of the plane pointed 22° west of north, you would be traveling at this speed in that direction if there were no wind. With the wind blowing, you are actually traveling due north at an effective speed of (750 km)/(4.32 h) = 173.6 km/h. The wind must therefore have a north-south component of (238.8 km/h)(cos 22°) − 173.6 km/h = 47.8 km/h (toward the south) and an east-west component of (238.8 km/h)(sin 22°) = 89.5 km/h (toward the east). The wind speed is thus v=
(47.8 km/h) 2 + (89.5 km/h) 2 = 101 km/h
in a direction that makes an angle of
θ= tan −1
2. (a)
89.5 km/h = 62° east of south 47.8 km/h
95 m = 179 s 0.53 m/s
(b)
95 m = 54 s 1.24 m/s + 0.53 m/s
(c)
95 m = 49 s 2.48 m/s − 0.53 m/s
3. ∆t = tup + tdown − 2tacross =
2L 1 1 − 2 2 c 1 − u / c 1 − u 2 / c2
Assuming u c , u2 1 ≅ + 1 c2 1 − u 2 / c2
Chapter 2
and
1 1 − u 2 / c2
∆t ≅
2 L u 2 1 u 2 Lu 2 1 + − 1 + = c c 2 2 c 2 c3
u=
c3 ∆t = L
≅ 1+
1 u2 2 c2
(3 ×108 m/s)3 (2 ×10−15 s) = 7 ×104 m/s 11 m
Page 1
4. (a)
u = 100 km/h = 28 m/s c 1 − u 2 / c2 ≅ 1 −
1 u2 1 (28 m/s) 2 = 1 − = 1 − 4.3 ×10−15 2 c2 2 (3 ×108 m/s) 2
L = L0 1 − u 2 / c 2 = L0 (1 − 4.3 ×10−15 ) L0 − L= (4.3 ×10−15 ) L0= (4.3 ×10−15 )(4 ×106 m)= 1.7 ×10−8 m This is less than the wavelength of light. (b) = ∆t
∆t0 1− u / c 2
2
≅
∆t0 ≅ ∆t0 (1 + 4.3 ×10−15 ) −15 1 − 4.3 ×10
∆t − ∆t0= (4.3 ×10−15 )∆t0= (4.3 × 10−15 )(40 h)(3600 s/h)= 6.2 ×10−10 s 1 5. With L = 12 L0 , the length contraction formula gives = L0 1 − u 2 / c 2 , so 2 L0
= u
3 / 4= c 2.6 ×108 m/s
6. The astronaut must travel 600 light-years at a speed close to the speed of light and must age only 12 years. To an Earth-bound observer, the trip takes about Δt = 600 years, but this is a dilated time interval; in the astronaut’s frame of reference, the elapsed time is the 2 2 proper time interval Δt 0 of 12 years. Thus, with ∆t =∆t0 / 1 − u / c ,
600 years =
12 years u 2 12 or 1− 2 = c 600 1 − u 2 / c2
2
u= 1 − (12 / 600) 2 c = 0.9998c
7. (a) = ∆t
∆t0 = 1 − u 2 / c2
120.0 ns = 384 ns 1 − (0.950) 2
(b)
d = v ∆t = 0.950(3.00 ×108 m/s)(384 ×10−9 s) = 109 m
(c)
d 0 = v ∆t0 = 0.950(3.00 ×108 m/s)(120.0 ×10−9 s) = 34.2 m
Chapter 2
Page 2
8. In the laboratory reference frame, the lifetime is d 1.15 mm = = 0.386 ×10−11 s u 0.993(3.00 ×108 m/s)
∆t =
∆t0 =∆t 1 − u 2 / c 2 =(0.386 ×10−11 s) 1 − (0.993) 2 =4.56 ×10−13 s
9. From Equation 2.15, ∆= t1 L /(v − u ) , and from Equation 2.16, ∆t= L /(c + u ) . 2 ∆t = ∆t1 + ∆t2 =
∆t0 L L + = v−u c+u 1 − u 2 / c2
With ∆t= L0 / v′ + L0 / c and = L L0 1 − u 2 / c 2 , this becomes 0 1 1 1 1 1 += + 2 2 v − u c + u 1 − u / c v′ c
Solving for v, we obtain v=
10.
v′ + u 1 + v′u / c 2
Let ship A represent observer O, and let observer O′ be on Earth. Then v′ = 0.831c and u = −0.743c, and so = v
v′ + u 0.831c + 0.743c = = 0.973c 2 1 + v′u / c 1 + (0.831)(0.743)
If now ship B represents observer O, then v′ = −0.743c and u = −0.831c. v= 11.
v′ + u −0.743c − 0.831c = = −0.973c 2 1 + v′u / c 1 + (−0.743)(−0.831)
Let O′ be the observer on the space station, and let O be the observer on ship B. Then v′ = 0.811c and u = −0.665c. = v
Chapter 2
v′ + u 0.811c − 0.665c = = 0.317c 2 1 + v′u / c 1 + (0.811)(−0.665)
Page 3
12.
(a) With f ′ =f (1 − u / c) /(1 + u / c) and λ = c / f , we obtain
1+ u / c 1+ u / c = or 366 nm 122 nm λ ′ λ= 1− u / c 1− u / c Solving, we get u/c = 0.800 or u = 2.40 × 108 m/s.
1− u / c 1 − 0.800 = (b) λ ′ λ= 122 nm = 40.7 nm 1+ u / c 1 + 0.800
13.
With f ′ =f (1 − u / c) /(1 + u / c) and λ = c / f , we obtain 2
2
1 − u / c f ′ λ 650 nm = = = 1.397 = 1 + u / c f λ ′ 550 nm 2
Solving, u/c = 0.166 or u = 5. 0 × 107 m/s. 14.
According to O, the rest length L 0 of the hypotenuse is shortened to L =L0 1 − u 2 / c 2 =L0 1 − (0.92) 2 =0.392 L0
The height of the triangle, which is 0.5L 0 , is not affected by the length contraction. The base angles of the triangle are then 0.5 L0 −1 = = = 68.6° θ tan −1 1 tan (2.55) × L 0.392 0 2
The apex angle is then 180° − 2(68.6°) = 42.8°. 15.
In the direction of motion, the radius of the gold nucleus is reduced to (7.0 fm) 1 − (0.99995) 2 = 0.070 fm
At this speed the formerly spherical gold nucleus looks like a pancake whose thickness is only 1% that of the original sphere and thus whose density has been increased by a factor of 100. The collision of two such compressed, high-energy nuclei is thought to create for a fraction of a second the conditions of density and temperature that occurred in the very early universe.
Chapter 2
Page 4
16. = dx′
= v′x
dx − u dt dt − u dx / c 2 = and dt ′ 1 − u 2 / c2 1 − u 2 / c2 vx − u dx′ dx − u dt dx / dt − u = = = 2 2 dt ′ dt − u dx / c 1 − u (dx / dt ) / c 1 − uvx / c 2
With dz ′ = dz , we obtain v′z =
17.
dz ′ dz (dz / dt ) 1 − u 2 / c 2 vz 1 − u 2 / c 2 = = = dt ′ (dt − u dx / c 2 ) / 1 − u 2 / c 2 1 − uvx / c 2 1 − uvx / c 2
For the light beam, observer O measures v x = 0, v y = c. Observer O′ measures vy 1 − u 2 / c2 v −u ′ = − = − = =c 1 − u 2 / c 2 v′x = x 0 u u and v y 2 2 1 − uvx / c 1 − uvx / c
According to O′, the speed of the light beam is v′ = 18.
(v′x ) 2 + (v′y ) 2 = u 2 + c 2 (1 − u 2 / c 2 ) = c
O measures times t 1 and t 2 for the beginning and end of the interval, while O′ measures t1′ and t2′ . Using Equation 2.23d, = t1′
t1 − ux / c 2 = and t2′ 2 2 1− u / c
t2 − ux / c 2 1 − u 2 / c2
The same coordinate x appears in both expressions, because the bulb is at rest according to O (so ∆t is the proper time interval). Subtracting these two equations, we obtain = t2′ − t1′
19.
t2 − t1 = ∆t ′ or 1 − u 2 / c2
∆t 1 − u 2 / c2
Suppose observer O is moving with the K meson; to this observer, the K meson appears to be at rest, and so O measures v 1 = +0.815c and v 2 = −0.815c for the two p mesons. Observer O′ is moving relative to O with a velocity u = −0.453c; in the reference frame of O′, observer O and the K meson are moving in the positive x direction with a velocity of 0.453c. We can use the Lorentz velocity transformation (Equation 2.28a) to find the velocities of the two p mesons according to O′: Chapter 2
Page 5
20.
v1′ =
v1 − u +0.815c − (−0.453c) = = +0.926c 2 1 − v1u / c 1 − (0.815)(−0.453)
v2′ =
v2 − u −0.815c − (−0.453c) = = −0.575c 2 1 − v2u / c 1 − (−0.815)(−0.453)
Imagine the rod to be the hypotenuse of a right triangle having sides L x along the x axis and L y in the y direction. According to O′, the length L x is shortened by the length contraction, but the length L y is unaffected because it is perpendicular to the direction of motion. For= O, Ly Lx tan 34° , while for= O′, L′y Lx′ tan 52° where= Lx′ Lx 1 − u 2 / c 2 . Because Ly = L′y , we have Lx tan 34 = ° Lx′ tan 52 = ° Lx 1 − u 2 / c 2 tan 52°
or u= c 1 − (tan 34°) 2 / (tan 52°) 2 = 0.850c
21.
(a) Changing the coordinates in Equation 2.23d to intervals, we have ∆t − u ∆x / c 2 0.528 m s − (0.685)(49.5 m) / (300 m/m s) ′ ∆t = = =+0.570 m s 1 − u 2 / c2 1 − (0.685) 2 (b) Changing coordinates to intervals in Equation 2.23a, ∆x′ =
49.5 m − (0.685 × 300 m/m s)(0.528 m s) = =−81.0 m 1 − u 2 / c2 1 − (0.685) 2
∆x − u ∆t
The negative sign of ∆x′ indicates that O′ finds the two events in inverted locations compared with O; for example, if O finds that event 1 occurs at a smaller x coordinate than event 2, then O′ finds that event 1 occurs at a larger x′ coordinate than event 2. That is, O sees event 1 to the left of event 2, while O′ sees event 1 to the right of event 2. Note than both observers find the time interval to be positive – event 2 occurs after event 1 to both observers.
22.
From Equation 2.23d written in terms of intervals, for O′ to find ∆t′ = 0, it must be true that ∆t − (u / c 2 )∆x =0 . Thus
Chapter 2
Page 6
u= 23.
c 2 ∆t (300 m/m s)(0.138 m s − 0.124 m s) = c = +0.32c ∆x (23.6 m − 10.4 m)
(a) In your frame of reference on the plane, the distance between Los Angeles and Boston is contracted to (3000 mi) 1 − [(600 mi/h)/(1000 mi/h)]2 = 2400 mi , and you measure the time for the trip to be (2400 mi)/(600 mi/h) = 4 h. So your watch reads 2:00 pm. (b) According to an observer on the ground, the trip takes (3000 mi)/(600 mi/h) = 5 h. So the airport clock reads 3:00 pm. (c) The return trip again takes 4 h as measured in your frame of reference and 5 h from the frame of reference of the ground. When you depart Boston, the airport clock reads 10:00 am but your watch reads 9:00 am, so when you land in Los Angeles, your watch reads 1:00 pm and the airport clock reads 3:00 pm. This analysis duplicates the result of the twin paradox: If your twin wearing an identical watch had remained at the Los Angeles airport, all observers would agree that you were 2 hours younger than your twin.
24.
(a) On the outward journey at 0.60c, the rate at which signals are received is
1− u / c 1 − 0.60 = f ′ f= (1/ year)= 0.5 / year 1+ u / c 1 + 0.60 (b) During the return journey,
1+ u / c 1 + 0.60 = f ′ f= (1/ year)= 2 / year 1− u / c 1 − 0.60 (c) According to Casper, Amelia’s outward journey lasts 16 years (it is 16 years before he sees her arrive at the planet), during which he receives 8 signals (0.5/year × 16 years). Her total journey lasts 20 Earth years, so the return journey lasts 4 Earth years, during which he receives 8 signals (2/year × 4 years). Thus Casper receives 16 signals (8 during the outward trip and 8 during the return) and concludes that his sister has aged 16 years.
25.
According to Amelia, the distance to the star is shortened to L =L0 1 - v 2 / c 2 =(8.0 l-y) 1 - (0.80) 2 =4.8 l-y
and at a speed of 0.80c Amelia’s travel time to the star is (4.8 l-y)/(0.80c) = 6.0 y. The total round-trip time in Amelia’s frame of reference is 12 years, so she is 8 years younger than her brother when she returns.
Chapter 2
Page 7
26.
(a) 20
15 Time (years) 10
5 Casper’s worldline
Amelia’s worldline
0
5
10
Distance (light-years)
(b) 16 years (c) 4 years 27.
(a) To an Earth-bound observer Alice’s round trip takes 20 years each way (20 years × 0.6c = 12 light-years) for a total time of 40 years. Bob’s travel time is 15 years each way (15 years × 0.8c = 12 light-years) for a total travel time of 30 years. With Bob’s 10-year delay in departing, the two arrive on Earth simultaneously. (b) To Alice, the distance to the star is contracted to L =L0 1 - v 2 / c 2 =12 light-years 1 - (0.6) 2 =9.6 light-years
So in Alice’s frame of reference the trip takes a time of (9.6 light years)/0.6c = 16 years each way. To Bob, the distance to the star is L =L0 1 - v 2 / c 2 =12 light-years 1 - (0.8) 2 =7.2 light-years
Chapter 2
Page 8
and in Bob’s frame the travel time is (7.2 light-years)/0.8c = 9 years each way. Relative to Alice’s original departure time, Alice has aged 32 years while Bob has aged 10 + 18 = 28 years. So Bob is younger by 4 years. 28.
(a) Suppose Agnes travels at speed v. Then in her reference frame the distance to the star is shortened to= L L0 1 − v 2 / c 2 , so the time for her one-way trip is L/v and thus 16 light-years 1 - v 2 / c 2 c2 10 = 10 y = or -1 2 v v 16
Solving, we find v = 0.848c. (b) According to Bert, Agnes traveled on a journey of 32 light-years at a speed of 0.848c which corresponds to a time of (32 light-years)/0.848c = 37.7 years. 29.
(a) m1c 2
K i′ = K1i′ + K 2i′ = (2m)c 2
=
1− 0
2
1 − v′ / c 2 1i
=
2
1 − v′ / c 2 2i
mc 2
− (2m)c 2 +
1 − (0.750)
m1c
K f′ = K1f′ + K 2f′ =
m2 c 2
− m1c 2 +
2
1 − v′ / c
2
− m2 c 2
2
2 − mc= 0.512mc 2
m2 c 2
− m1c 2 +
− m2 c 2
1 − v′ / c (2m)c mc 2 2 − (2m)c + = − mc 2 0.512mc 2 2 2 1 − (−0.585) 1 − (0.294) 2 1f
2
2 2f
2
2
(b) K i = K1i + K 2i =
m1c 2 1− v / c 2 1i
(2m)c 2
=
1 − (0.550)
2
30. = p
1− v / c 2 2i
2
− m2 c 2
= − mc 2 0.458mc 2
1 − (−0.340) 2
m1c 2 1− v / c
m2 c 2 mc 2
− (2m)c 2 +
K f = K1f + K 2f = =
2
− m1c 2 +
− m1c 2 +
m2 c 2
− m2 c 2
1− v / c (2m)c mc 2 2 − (2m)c + = − mc 2 0.458mc 2 2 2 1 − (−0.051) 1 − (0.727) 2 1f
2
2 2f
2
2
mv 1 (mc 2 )(v / c) 1 (938.3 MeV)(0.835) = = = 1424 MeV/c 1 − v2 / c2 c 1 − v2 / c2 c 1 − (0.835) 2
Chapter 2
Page 9
= K
mc 2 1− v / c 2
2
938.3 MeV
−= mc 2
1 − (0.835) 2
− 938.3 = MeV 767 MeV
E= K + mc 2 = 767 MeV + 938.3 MeV = 1705 MeV 31.
E= K + mc 2 = 0.923 MeV + 0.511 MeV = 1.434 MeV Solving Equation 2.36 for v, we obtain 2
2
mc 2 0.511 MeV v =− c 1 c 1 0.934c =− = 1.434 MeV E 32.
= W = ∫ F dx
dp
dx
dx ∫= dp v dp ∫= dt dt ∫
v v v mv 2 mv K= v dp = pv − p dv = −∫ dv ∫0 ∫0 0 1 − v2 / c2 1 − v2 / c2 mv 2 mc 2 = + mc 2 1 − v 2 / c 2 −= mc 2 − mc 2 2 2 2 2 1− v / c 1− v / c
33.
For what range of velocities is K − 12 mv 2 ≤ 0.01K ? At the upper limit of this range, where K − 12 mv 2 = 0.01K , we have
mc 2 = = − mc 2 0.99 K 0.99 2 2 1− v / c
1 2
mv 2
1 0.5 1 − 1 = 12 x which gives = 1 + x With x = v / c , 0.99 1 − x 0.99 1− x 2
2
2
1= (1 − x)(1 + 1.0101x + 0.2551x 2 )
or
0.2551x 2 + 0.7550 x − 0.0101 = 0
Solving using the quadratic formula, we find x = 0.0133 or -2.97. Only the positive solution is physically meaningful, so = v
= 0.0133 c 0.115c
That is, for speeds smaller than 0.115c, the classical kinetic energy is accurate to within 1%. For a different approach to that same type of calculation, see Problem 36.
34.
As in Problem 33, let us now find the lower limit on the momentum such that Chapter 2
Page 10
( pc) 2 + (mc 2 ) 2 − pc ≤ 0.01 ( pc) 2 + (mc 2 ) 2
From the lower limit, we obtain 0.99 ( pc) 2 + (mc 2 ) 2 = pc , which can be written as
= ( pc) 2
m2c 4 = or pc 7.02mc 2 1/(0.99) 2 − 1
7.02mc 2 , we obtain With mvc / 1 − v 2 / c 2 = v2 v2 = 49.25 1 − 2 c2 c
or
v/c = 0.990
Whenever v /c ≥ 0.990 , the expression E = pc will be accurate to within 1%. 35.
1 − v2 / c2 + v2 / c2 (mc 2 ) 2 m2c 2v 2 2 2 = + = E2 = 2 2 = (mc 2 ) 2 ( mc ) (mc 2 ) 2 + ( pc) 2 2 2 2 2 − − 1− v / c 1 v / c 1 v / c
= E
36.
(mc 2 ) 2 + ( pc) 2
With (1 + x) n =+ 1 nx + n(n − 1) x 2 /2! we have 2
1 v 2 (−1/ 2)(−3 / 2) v 2 = + 1+ 2 + 2 c2 2 1 − v2 / c2 c 1
and so
1 v2 3 v4 1 1 mc 2 1 + 1 K mc 2 = −= + 4 + −= 2 2 2 2c 8c 1− v / c
1 2
3 v2 mv 2 1 + + 2 4c
The correction term is 3v 2 /4c 2 , which has the value 0.1% when 3v 2 /4c 2 = 0.001 , or
= v
37.
= 0.001(4 / 3) c 0.0365c
(a) With E = 1351 MeV and pc = 1256 MeV, Equation 2.39 gives m=
Chapter 2
1 c2
E 2 − ( pc) 2 =
1 (1351 MeV) 2 − (1256 MeV) 2 = 498 MeV/c 2 2 c
Page 11
(b)
E = ( pc) 2 + (mc 2 ) 2 = (857 MeV) 2 + (498 MeV) 2 = 991 MeV
38. K f − K i = ( Ef − mc 2 ) − ( Ei − mc 2 ) = Ef − Ei =
mc 2 1− v / c 2 f
2
−
1− v / c 2 i
2
=
0.511 MeV 1 − (0.91)
2
−
0.511 MeV 1 − (0.85) 2
= 0.262 MeV
∆E = mc ∆T = (1 g)(0.40 J/g ⋅ K)(100 K) = 40 J
39.
∆m =
40.
mc 2
40 J ∆E = = 4.4 ×10−16 kg 2 16 2 2 9 ×10 m /s c
(a) At such low speed, the classical approximation is valid.
v2 K =12 mv 2 =12 mc 2 2 =12 (0.511 MeV)(1.00 ×10-4 ) 2 =2.56 ×10-3 eV c (b) The relativistic expression gives
1 1 = K mc 2 = − 1 0.511 MeV = − 1 25.6 eV 2 2 1 − (0.01) 2 1− v / c For this speed, the classical expression 12 mv 2 also gives 25.6 eV, so the two calculations agree to at least three significant figures. (Actually they agree to four significant figures, but not to five.) (c) The relativistic expression gives
1 1 = − 1 0.511 MeV = − 1 24.7 keV K mc 2 = 2 2 1 − (0.3) 2 1− v / c For this speed, the classical expression gives 23.0 keV, which is incorrect by about 7%.
1 1 = − 1 0.511 MeV = − 1 10.9 MeV (d) K mc 2 = 2 2 1 − (0.999) 2 1− v / c
Chapter 2
Page 12
41.
Because the electrons and the protons have charges of the same magnitude e, after acceleration through a potential difference of magnitude ∆V = 12.0 million volts (a positive difference for the electron, a negative difference for the proton), each loses potential energy of ∆U =−e ∆V = -12.0 MeV and thus each acquires a kinetic energy of K = +12.0 MeV. For the electron, E = K + mc2 = 12.0 MeV + 0.511 MeV = 12.5 MeV. The momentum is then p=
1 1 E 2 − (mc 2 ) 2 = (12.5 MeV) 2 − (0.511 MeV) 2 =12.5 MeV/c c c
The classical formula K = p 2 / 2m gives = p
= 2mK
2(0.511 MeV/c 2 )(12.0 MeV) = 3.50 MeV/c
which is far from the correct result (a discrepancy we would expect for such highly relativistic electrons). For the protons, E = K + mc2 = 12.0 MeV + 938.3 MeV= 950.3 MeV, and the momentum is p=
1 1 E 2 − (mc 2 ) 2 = (950.3 MeV) 2 − (938.3 MeV) 2 =150.5 MeV/c c c
The classical formula gives = p
= 2mK
2(938.3 MeV/c 2 )(12.0 MeV) = 150.1 MeV/c
The difference between the classical and relativistic formulas appears only in the fourth significant figure.
42.
The mass of a uranium atom is about (235 u)(1.66 × 10−27 kg/u) = 3.90 × 10−25 kg, so 1.50 kg contains 1.50 kg/3.90 × 10−25 kg = 3.84 × 1024 atoms. The total energy released is ∆E= (210 MeV/atom)(3.84 ×1024 atoms)= 8.06 ×1026 MeV and the change in mass is ∆m=
∆E (8.06 ×1026 MeV)(1.602 ×10−13 J/MeV) = = 1.44 ×10−3 kg c2 (2.998 ×108 MeV) 2
About one gram of matter vanishes for each kilogram that fissions! 43.
(a) The change in mass is
Chapter 2
Page 13
∆ = m m( 2 H) + m(n) − m(3 H) = 2.014102 u + 1.008665 u − 3.016049 u = 0.006718 u
The corresponding energy change is ∆E =∆m c 2 =(931.5 MeV/u)(0.006718 u) =6.26 MeV . (b) = ∆m m(3 He) + m(n) − m( 4 He) = 3.016029 u + 1.008665 u − 4.002603 u = 0.022352 u
∆E =∆m c 2 =(931.5 MeV/u)(0.022352 u) =20.8 MeV
44.
(a) E = Ep + Ep =
mp c 2 1 − v /c 2
2
+ mp c 2 =
139.6 MeV 1 − (0.921) 2
+ 938.3 MeV = 1296.7 MeV
(b) 1 mp c 2 (v /c) 1 (139.6 MeV)(0.921) p = pp + pp = = = = 330.0 MeV/c 1 − v 2 /c 2 c 1 − v 2 /c 2 c 1 − (0.921) 2 mp v
(c) mc 2 = E 2 − ( pc) 2 = (1296.7 MeV) 2 − (330.0 MeV) 2 =1254 MeV
45.
Before the collision, the total relativistic energy of each electron is = Ee
me c 2 = 1 − v2 / c2
0.511 MeV = 114.3 MeV 1 − (0.99999) 2
The total energy in the collision is therefore 2×114.3 MeV = 228.6 MeV. The total momentum is zero before the collision, because the two particles moves with equal and opposite velocities and have equal masses. After the collision, the total momentum is still zero, so we know that the two muons must move with equal speeds and thus have equal energies. The total energy of each muon is then 114.3 MeV and its kinetic energy is Km = Em − mm c 2 = 114.3 MeV − 105.7 MeV = 8.6 MeV 46.
The two protons have equal (and opposite) momenta and thus equal energies E 1 and E 2 . The new particle is created with zero momentum (at rest), so its total energy is equal to its rest energy Mc2 = 9460 MeV. Conservation of energy then gives E1 + E2 = Mc 2 , so E1 = E2 = Mc 2 /2 .
Chapter 2
Page 14
mp c 2 Mc 2 = 2 1 − v 2 /c 2
= E1
2
2 v 2 2(mp c ) 2(938.3 MeV) = = 1− 2 = 0.0394 2 c Mc 9460 MeV
47.
2
v 0.980c so=
For particle 1, moving in the positive x direction,
E1 =K1 + mc 2 =282 MeV + 140 MeV =422 MeV cp1 = E12 − (mc 2 ) 2 = (422 MeV) 2 − (140 MeV) 2 = +398 MeV For particle 2, moving in the negative x direction,
E2 =K 2 + mc 2 =25 MeV + 140 MeV =165 MeV cp2 = − E22 − (mc 2 ) 2 = − (165 MeV) 2 − (140 MeV) 2 = −87 MeV The net final momentum is pf = p1 + p2 = 398 MeV/c − 87 MeV/c = 311 MeV/c , and the net final energy is Ef = E1 + E2 = 422 MeV + 165 MeV = 587 MeV . Because of the conservation laws, these must be equal to the momentum and the energy of the initial particle, so that its rest energy is then
mi c 2 = Ei2 − (cpi ) 2 = (587 MeV) 2 − (311 MeV) 2 =498 MeV Solving Equation 2.36 for v, we obtain 2
2
mc 2 498 MeV v =− c 1 c 1 0.529c =− = 587 MeV E
48.
(a) In this frame of reference, the total momentum is zero before and after the collision, and at threshold the four product particles are formed at rest. So the total final energy is 4mc2, and since this must be equal to the total initial energy each initial colliding proton must have an energy of 2mc2. (b) With = E mc 2 / 1 − v 2 / c 2 , we have for each of the initial protons = 2mc 2
mc 2 1− v / c 2
2
= or 1 − v2 / c2
1 2
so v = 0.866c.
Chapter 2
Page 15
(c) With the transformation speed u = 0.866c, we have = v′
v−u −0.866c − 0.866c = = 0.9897c 2 1 − vu / c 1 − (−0.866)(0.866)
The proton energy in this frame is then = E′
49.
pc =
mc 2 = 1 − v ′2 / c 2
(mc 2 )(v / c)
v = c = ∆t
1 − v2 / c2 ( pc / mc 2 ) 2 = 1 + ( pc / mc 2 ) 2
(3094 MeV/105.7 MeV) 2 = 0.99942 1 + (3094 MeV/105.7 MeV) 2
2.198 m s = 64.38 m s 1- (0.99942) 2
∆t0 = 1 - v2 / c2
( pc) 2 =− E 2 (mc 2 ) 2 =+ ( K mc 2 ) 2 − (mc 2 ) 2 =+ K 2 2 Kmc 2
50.
K2 p = 2 Km + 2 so c 2
51.
mc 2 = 7 mc 2 1 − (0.9897) 2
p2 K = m+ 2 2K 2c
(a) If the astronaut travels at speed v for a distance d = 200 ly in the reference frame of Earth, then in the spacecraft reference frame the distance to the star is d 1 − v 2 / c 2 and the time T for the astronaut to reach the star (which must be 10 years in the spacecraft frame of reference) is
d 1 − v2 / c2 = T v
or
T v cT 1 − v2 / c2 = v = d c d
Solving, we find 1 1 v c= c = 0.99875c = 2 1 + (Tc / d ) 1 + (10 / 200) 2
(b) According to an observer on Earth, the astronaut travels a total distance of 400 ly at a speed of 0.99875c, so the total time for the round trip is (400 ly)/0.99875c = 400.5 y.
Chapter 2
Page 16
52. = t1′
t1 − (u / c 2 ) x1 t − (u / c 2 ) x2 = and t2′ 2 1 − u 2 / c2 1 − u 2 / c2 = t2′ − t1′
t2 − t1 1 − u 2 / c2
−
(u / c 2 )( x2 − x1 ) 1 − u 2 / c2
In the O frame of reference, the “cause” can travel to the “effect” with a speed that can be no greater than c; that is, ( x2 − x1 ) /(t2 − t1 ) ≤ c , or ( x2 − x1 ) ≤ c(t2 − t1 ) . Substituting for ( x2 − x1 ) in the second term, we obtain t2 − t1
t2′ − t1′ ≥
1 − u 2 / c2
−
(u / c)(t2 − t1 ) 1 − u 2 / c2
= (t2 − t1 )
1− u / c 1 − u 2 / c2
≥0
For the red flash, = x1 0= at t1 0, = so x1′ 0 and = t1′ 0. For the blue flash, which occurs at = x2 3.65 = km and t2 8.24 m s, 53.
x2′ =
x2 − ut2 1− u / c 2
2
=
3.65 km − (0.534c)(8.24 m s) 1 − (0.534) 2
= −0.89 km
t2 − (u / c 2 ) x2 8.24 m s − (0.534 / 0.300 km/m s)(3.65 km) = = 2.06 m s t2′ = 1 − u 2 / c2 1 − (0.534) 2
54.
Let O′ be the observer on ship A, traveling at u = 0.65c relative to the space station. Observer O on the space station measures v Bx = 0, v By = 0.50c for ship B. According to O′, vBx − u 0 − 0.65c ′ = = = −0.65c vBx 2 1 − vBx u / c 1− 0 ′ = vBy
vBy 1 − u 2 / c 2 0.50c 1 − (0.65) 2 = = 0.38c 1 − vBx u / c 2 1− 0
′ ) 2 + (vBy ′ )2 = vB′ = (vBx (−0.65c) 2 + (0.38c) 2 = 0.75c v′ ′ vBx
0.38c = 150° −0.65c For ship C, O measures v Cx = −0.50c, v Cy = 0. By tan −1 = tan −1 θ= B
Chapter 2
Page 17
−0.50c − 0.65c ′ = ′ = vCx = −0.87c, vCy 0 1 − (−0.50)(0.65) ′ 0.87c at θ= v= 180° (negative x direction) C C For ship D, O measures vDx = (−0.50c) sin 45° = −0.35c, vDy = +0.35c . 0.35c 1 − (0.65) 2 −0.35c − 0.65c ′ = ′ = vDx = −0.81c, vDy = 0.22c 1 − (−0.35)(0.65) 1 − (−0.35)(0.65) vD′ = (−0.81c) 2 + (0.22c) 2 = 0.84c at θ D = tan −1
0.22c =° 165 −0.81c
55. (a) t ′ =
t − (u / c 2 ) x 1.52 m s − (0.563 / 300 m/m s)(524 m) = = 0.648 m s 1 − u 2 / c2 1 − (0.563) 2
(b) x′ =
x − ut 524 m − (0.563 × 300 m/m s)(1.52 m s) = = 335 m 2 2 1− u / c 1 − (0.563) 2
56.
(a) O′ measures v′ =− (v u ) /(1 − uv / c 2 ), and according to O′ the energy is = E′
mc 2 = 1 − v′2 / c 2
mc 2 2
1 − (v − u ) /(1 − uv / c 2 / c 2
and the momentum is = p′
mv′ = 1 − v′2 / c 2
m(v − u ) /(1 − uv / c 2 ) 2
1 − (v − u ) /(1 − uv / c 2 ) / c 2 2
m 2 c 4 − m 2 (v − u ) /(1 − uv / c 2 ) 2 2 (b)= E ′ − ( p′c) = m2c 4 2 2 2 1 − (v − u ) /(1 − uv / c ) / c The quantity E ′2 − ( p′c) 2 is equal to m2c4 in every frame of reference, no matter what its relative speed u. In other words, every observer measures the same value for the rest energy or mass. 57.
−u , v′y =− v 1 u 2 / c2 . (a) O′ measures v′x =
Chapter 2
Page 18
v′=
(v′x ) 2 + (v′y ) 2 =
= E′
mc 2 = 1 − v′2 / c 2
= p′
mv′ = 1 − v′2 / c 2
(b) = E ′2 − ( p′c) 2
u 2 + v 2 − u 2v 2 / c 2 mc 2 1 − u 2 / c 2 − v 2 / c 2 + u 2v 2 / c 4 m u 2 + v 2 − u 2v 2 / c 2 1 − u 2 / c 2 − v 2 / c 2 + u 2v 2 / c 4
m 2 c 4 − m 2 c 2 (u 2 + v 2 − u 2 v 2 / c 2 ) = m2c 4 2 2 2 2 2 2 4 1− u / c − v / c + u v / c
58. 40 Time (years) 30 Amelia’s return worldline
20 20
16
Bernice’s worldline 10 Casper’s worldline
-20
-10 Distance (light-years)
Amelia’s outbound worldline
4 0
(b) 40 years (c) During Amelia’s outbound journey, her worldline and Bernice’s overlap. This part of her journey takes 8 years, as measured in the clocks in Bernice’s frame of reference. (d) Let O = Casper and O′ = Bernice. Then u = 0.60c and v x = −0.60c.
Chapter 2
Page 19
vx − u −0.60c − 0.60c = = −0.882c 2 1 − vx u / c 1 − (−0.60)(0.60)
v′x =
(e) On his 4th birthday. (f) On his 16th birthday.
59.
(a) Before the first acceleration, E = E 0 = mc2. After the acceleration, the energy is = E1
mc 2 = 1 − v2 / c2
0.511 MeV = 3.6 MeV 1 − (0.99) 2
The change in energy is ∆E = E1 − E0 = 3.6 MeV − 0.5 MeV = 3.1 MeV , so the first stage adds 3.1 MeV to the energy of the electron. (b) = E2
mc 2 = 1 − v2 / c2
0.511 MeV = 11.4 MeV 1 − (0.999) 2
The change in energy is ∆E = E2 − E1 = 11.4 MeV − 3.6 MeV = 7.8 MeV , so the second stage adds about 2.5 times as much energy as the first stage, even though the second stage increases the velocity by only 0.9%.
60.= K
mc 2 1− v / c 2
2
−= mc 2
0.511 MeV 1 − (0.813) 2
− 0.511 = MeV 0.367 MeV per particle
Ebeam = (0.367 MeV/particle)(2.14 ×1011 particles/s)(3600 s) =× 2.82 1014 MeV = 45.2 J
Copper has a density of ρ = 8.92 g/cm3 and a specific heat capacity of c p = 0.385 J/g⋅K. 3 The mass M of the copper is = = M ρ= V (8.92 g/cm3 )(2.54 cm) 146 g . Assuming all of the kinetic energy carried by the particles in the beam acts to produce a change in temperature of the copper ( Ebeam = Mcp ∆T ), the temperature increase is
∆= T
61. = (a) pi
Chapter 2
Ebeam 45.2 J = = 0.80 = K 0.80 C° Mcp (146 g)(0.385 J/g ⋅ K)
mvi (0.511 MeV/c 2 )(0.960c) = = 1.752 MeV/c 1 − vi2 / c 2 1 − (0.960) 2
Page 20
= p1f
mv1f (0.511 MeV / c 2 )(0.956c) = = 1.665 MeV/c 1 − v1f2 / c 2 1 − (0.956) 2
pxf = p1f cos θ1 + p2f cos θ 2 =(1.665 MeV/c)(cos 9.7°) + p2f cos θ 2 =1.642 MeV/c + p2f cos θ 2 p yf = p1f sin θ1 + p2f sin θ 2 =(1.665 MeV/c)(sin 9.7°) + p2f sin θ 2 =0.281 MeV/c + p2f sin θ 2 Conservation of momentum gives p xf = p i and p yf = 0. Thus p2f cos θ 2 = 1.752 MeV/c − 1.642 MeV/c = 0.110 MeV/c −0.281 MeV/c p2f sin θ 2 = 0 − 0.281 MeV/c = Dividing the second result by the first gives −0.281 MeV/c tan θ 2 = = −2.55 0.110 MeV/c
or
−68.6° θ2 =
With p2f (0.110 MeV/= = c) /[cos(−68.6°)] 0.302 MeV/c , we have mv2f = p2f 0.302 = MeV/c 1 − v2f2 / c 2 and solving, we find v 2f = 0.508c. (b)
= Ei
mc 2 = 1 − vi2 / c 2
= E1f
0.511 MeV = 2.336 MeV 1 − (0.960) 2
mc 2 = 1 − v1f2 / c 2
0.511 MeV = 1.743 MeV 1 − (0.956) 2
Conservation of energy gives E i = E 1f + E 2f , so E 2f = 2.366 MeV − 1.743 MeV = 0.593 MeV and mc 2 0.511 MeV MeV = E2f 0.593 = = 1 − v2f2 / c 2 1 − v2f2 / c 2 and solving we find v 2f = 0.508c, in agreement with part (a). 62.
The initial energy is Eπ =
Chapter 2
mc 2 = 1 − v2 / c2
135 MeV = 678 MeV 1 − (0.98) 2 Page 21
and the momentum is pp =
1 1 Ep2 − (mc 2 ) 2 = (678 MeV) 2 − (135 MeV) 2 =664 MeV/c c c
Because the two gamma ray photons have equal energies, each has an energy of 1 2 (678 MeV) , so E γ = 339 MeV. Each gamma ray photon has a momentum of = pγ E= 339 MeV/c , which has a component pγ cos θ along the direction of the γ /c initial p meson. Conservation of momentum then gives pp = 2 pγ cos θ , so the angle is
pp 664 MeV/c cos −1 = θ cos −1 = = 11.7° 2 pγ 2(339 MeV/c) 63.
(a) The total energy of the K meson in this frame is E = mc2 + K = 497.7 MeV + 77.0 MeV = 574.7 MeV, and from Eq. 2.36 we then have mc 2 497.7 MeV 1 − v2 / c2 = = = 0.8660 E 574.7 MeV
from which v = 0.5000c. (b) In the original frame the energy of the pi meson is E = ( pc) 2 + (mc 2 ) 2 = (381.6 MeV) 2 + (139.6 MeV) 2 = 406.3 MeV
and we can find its speed as we did in part (a) mc 2 139.6 MeV 1 − v2 / c2 = = = 0.3436 E 406.3 MeV
so v = 0.9391c. Transforming this speed to the new reference frame gives = v′
v−u 0.9391c − 0.5000c = = 0.8278c 2 1 − uv / c 1 − (0.9391)(0.5000)
from which we find = p′ = E′
Chapter 2
1 mc 2 (v / c) 1 (139.6 MeV)(0.8278) = = 206.0 MeV/c c 1 − v2 / c2 c 1 − (0.8278) 2 mc 2 = 1 − v2 / c2
139.6 MeV = 248.9 MeV 1 − (0.8278) 2
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(c) Applying the conservation laws in this reference frame then gives -206.0 MeV/c for the momentum of the second particle and 497.7 MeV – 248.9 MeV = 248.8 MeV for its energy. Its mass is then, from Eq. 2.39, mc 2 = E 2 − ( pc) 2 = (248.8 MeV) 2 − (206.0 MeV) 2 =139.5 MeV
again showing that the second particle is another pi meson. 64.
(a) We use subscripts − and + to represent respectively the electron (e−) and positron (e+). Then before the collision the momenta are = p− p+ =
mv− 1 mc 2 (v− / c) 1 (0.511 MeV)(0.834) = = = 0.772 MeV/c 1 − v−2 / c 2 c 1 − v−2 / c 2 c 1 − (0.834) 2 mv+ 1 − v+2 / c 2
=
1 mc 2 (v+ / c) 1 (0.511 MeV)(−0.428) = = −0.242 MeV/c c 1 − v+2 / c 2 c 1 − (0.428) 2
The total momentum before the collision is p = p− + p+ = 0.772 MeV/c − 0.242 MeV/c = 0.530 MeV/c .
By conservation of momentum, this must also be the momentum of the new particle after the collision. The total energies before the collision are = E−
mc 2 = 1 − v−2 / c 2
0.511 MeV = 0.926 MeV 1 − (0.834) 2
= E+
mc 2 = 1 − v+2 / c 2
0.511 MeV = 0.565 MeV 1 − (0.428) 2
The total energy before the collision is E = E− + E+ = 0.926 MeV + 0.565 MeV = 1.492 MeV .
By conservation of energy, this must be the same as the total energy of the new particle after the collision. (b) We can find the mass of the new particle from its momentum and total energy using Eq. 2.39: M = c −2 E 2 − ( pc) 2 = c −2 (1.492 MeV) 2 − (0.530 MeV) 2 = 1.394 MeV/c 2
(c) The initial and final kinetic energies are
Chapter 2
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K i = ( E− − mc 2 ) + ( E+ − mc 2 ) = 1.492 MeV − 2(0.511 MeV) = 0.470 MeV K f =− E Mc 2 = 1.492 MeV − 1.394 MeV = 0.097 MeV
The change in kinetic energy is ∆K = K i − K f = 0.470 MeV − 0.097 MeV = 0.372 MeV
The change in mass is ∆m = M − 2m = 1.394 MeV/c 2 − 2(0.511 MeV/c 2 ) = 0.372 MeV/c 2
The additional mass of the new particle comes from the loss in kinetic energy in the collision. (d) The momentum and energy of the original particles and the new particle would have different values in the new frame, but the values for M, ∆m, and ∆K would be the same. Mass is an invariant in special relativity (all observers measure the same value) and since the mass value comes from the change in kinetic energy, all observers must also find the same ∆K.
Chapter 2
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Chapter 3
1.
Dy= yn +1 − y= n
2.
sin= θ
3. (a)= λ (b)
λD d
=
(589.0 nm)(2.604 m) = 1.23 mm 1.25 mm
nλ 2(0.250 nm) = = 0.8865 2d 2(0.282 nm)
so = θ sin −1 0.8865 = 62.4°
2d sin θ 2(0.327 nm)(sin 38.0°) = = 0.403 nm n 1
The spacing between planes is
2 d sin 45° (0.327 nm) 0.231 nm = = 2 0.403 nm λ sin= or θ = = 0.871 2d 2(0.231 nm)
θ 60.6° =
This is the angle that the reflected ray makes with the plane of atoms. Because this plane makes an angle of 45° with the surface, the beam emerges at an angle of 60.6° − 45 = ° 15.6° with the surface. 4. The Bragg formula for the first-order peak is λ = 2d sin θ . We need to know the relationship between a small angular range dθ and the corresponding wavelength range dλ, which is found by taking the differentials from the Bragg formula: d λ = 2d cos θ dθ
Eliminating the lattice spacing d using the Bragg formula, we obtain (after converting dθ to radians) λ = d λ 2= cos θ dθ λ cot θ dθ 2sin θ =(0.149 nm)(cot 15.15°)(2.6 ×10-4 rad) =1.4 ×10-4 nm 5. (a)
= = 1.60 × 10−12 J E 10.0 MeV
E 10.0 MeV = = 1.00 × 107 eV/c c c 1.60 × 10−12 J p= 5.33 × 10−21 kg ⋅ m/s = 8 3.00 × 10 m/s
p =
(b) = E 25 keV = 4.0 × 10−15 J Chapter 3
Page 1
E 25 keV = = 2.5 × 104 eV/c c c 4.0 × 10−15 J = p= 1.3 × 10−23 kg ⋅ m/s 8 3.00 × 10 m/s
= p
(c) = λ 1.0 m m = 1.0 × 103 nm p=
h 1 hc 1 1240 eV ⋅ nm = = = 1.2 eV/c λ c λ c 1.0 × 103 nm
6.6 × 10−34 J ⋅ s = p= 6.6 × 10−28 kg ⋅ m/s −6 1.0 × 10 m
(d) E = hf = (4.14 ×10−15 eV ⋅ s)(150 ×106 Hz) = 6.2 ×10−7 eV = 9.9 ×10−26 J
E 6.2 × 10−7 eV = = 6.2 × 10−7 eV/c c c −26 9.9 × 10 J p= = 3.3 × 10−34 kg ⋅ m/s 8 3.00 × 10 m/s
= p
15 6. At 1 MHz = 106 Hz, E = hf = (4.14 ×10-eV ⋅ s)(106 s -1 ) = 4 ×10 9 eV 15 At 100 MHz = 108 Hz, E = hf = (4.14 ×10-eV ⋅ s)(108 s -1 ) = 4 ×10 7 eV
The range is from 4 × 10−9 eV to 4 × 10−7 eV.
λ =
7. (a)
(b)= λ
hc 1240 eV ⋅ nm = = 0.124 nm E 1.00 × 104 eV
1240 eV ⋅ nm = 1.24 × 10−3 nm 6 1.00 × 10 eV
(c) 350 nm:
= E
700 nm: E =
hc 1240 eV ⋅ nm = = 3.5 eV λ 350 nm 1240 eV ⋅ nm = 1.8 eV 700 nm
The range is from 1.8 eV to 3.5 eV.
Chapter 3
Page 2