Download Chapter 2 Relativity II. Home Work Solutions

Chapter 2
Relativity II. Home Work Solutions
2.1
Problem 2.4 (In the text book)
A charged particle moves along a straight line in a uniform electric field E with a speed v.
If the motion and the electric field are both in the x direction,
(a) show that the magnitude of the acceleration of the charge q is given by
qE
dv
=
a=
dt
m
v2
1− 2
c
3/2
(b) Discuss the significance of the dependence of the acceleration on the speed.
(c) If the particle starts from rest at x = 0 at t = 0, find the speed of the particle and its
position after a time t has elapsed. Comment on the limiting values of v and x as t → ∞
Solution
(a) The electric force on the charged particle is
F = qE
dp
=
dt
d
=
(γmv)
dt
!
d
v
p
= m
dt
1 − v 2 /c2
2
CHAPTER 2. RELATIVITY II. HOME WORK SOLUTIONS
#
dv
1 −2v 2 /c2
= m p
−
1 − v 2 /c2 2 (1 − v 2 /c2 )3/2 dt
1 − v 2 /c2
dv
v 2 /c2
= m
+
2
2
3/2
2
2
3/2
(1 − v /c )
(1 − v /c )
dt
m
dv
=
(1 − v 2 /c2 )3/2 dt
"
1
since dv/dt is the acceleration a we get:
ma
(1 − v 2 /c2 )3/2
3/2
v2
qE
1− 2
a =
m
c
qE =
(b) The larger the speed the smaller the acceleration, and eventually when v → c, a → 0
and this makes sense since it is not possible to accelerate an object to speeds beyond c.
When v c then a ≈ qE/m which is the classical expression for the acceleration.
(c) Using the expression for the acceleration from section (a), and a = dv/dt, we get:
qE
dv
=
dt
m
dv =
cd(v/c)
=
(1 − v 2 /c2 )3/2
Z v
d(v/c)
=
2 2 3/2
0 (1 − v /c )
#v
"
(v/c)
p
=
1 − v 2 /c2 0
v2
1− 2
c
3/2
3/2
qE
v2
1− 2
dt
m
c
qE
dt
m
Z t
qE
dt
0 mc
t
qEt
mc 0
v/c
qEt
p
=
mc
1 − v 2 /c2
2
v 2 /c2
qEt
=
1 − v 2 /c2
mc
2
2
v
v2
qEt
=
1− 2
c2
c
mc
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.1. PROBLEM 2.4 (IN THE TEXT BOOK)
3
"
2 #
2
v2
qEt
qEt
1+
=
c2
mc
mc
v2
(qEt/mc)2
=
c2
1 + (qEt/mc)2
1
=
1 + (mc/qEt)2
From the last equation we can see that when t → 0, v → 0 and when t → ∞, v → c.
using v = dx/dt we get:
dx
= v
dt
s
= c
= cp
dx = c p
=
Z
x
dx =
0
[x]x0 =
x =
(qEt/mc)2
1 + (qEt/mc)2
qEt
(mc)2 + (qEt)2
qEt dt
(mc)2 + (qEt)2
(qEt) d(qEt)
c
p
qE (mc)2 + (qEt)2
Z t
(qEt) d(qEt)
c
p
qE 0
(mc)2 + (qEt)2
h
it
c p
(mc)2 + (qEt)2
qE
0
i
c hp
(mc)2 + (qEt)2 − mc
qE
The last equation shows that as t → 0, x → 0 and as t → ∞, x → ∞
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
4
2.2
CHAPTER 2. RELATIVITY II. HOME WORK SOLUTIONS
Problem 2.14 (In the text book)
How long will the Sun shine, Nellie? The Sun radiates about 4.0 × 1026 J of energy into
space each second.
(a) How much mass is released as radiation each second?
(b) If the mass of the Sun is 2.0 × 1030 kg, how long can the Sun survive if the energy release
continues at the present rate?
Solution
(a) Since E = mc2 , the radiation released from the was produced by changing part of the
sun’s mass into energy, the equivalent mass released every second is then:
E
c2
4.0 × 1026
=
9 × 1016
= 4.4 × 109 kg
m =
(b) The life time of the sun if it keeps radiating at the current rat is:
2.0 × 1030
4.4 × 109
= 4.5 × 102 s
4.5 × 1020
=
years
60 × 60 × 24 × 365
= 1.4 × 1013 years
t =
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.3. PROBLEM 2.18 (IN THE TEXT BOOK)
2.3
5
Problem 2.18 (In the text book)
−
−
25
Consider the decay 24
55 Cr −→ 55 Mn + e , where e is an electron. The
mass of 54.9279 u, and the 55 Mn nucleus has a mass of 54.9244 u.
55
Cr nucleus has a
(a) Calculate the mass difference in M eV .
(b) What is the maximum kinetic energy of the emitted electron?
Solution
(a) The mass difference ∆m in atomic mass units u is:
∆m = m55 Cr − m55 Mn
= 54.9279 − 54.9244
= 3.5 × 10−3 u
Given that 1u = 931.5 M eV /c2 , then the mass difference in M eV is:
∆m = 3.5 × 10−3 × 931.5
= 3.26 M eV /c2
and the equivalent energy ∆E is:
∆E = 3.26 M eV
(b) The electron needs to be created, i.e. its rest mass energy has to come from somewhere.
∆E the total amount of energy available to create the electron and as kinetic energy for
the electron and 55 Mn. The maximum kinetic energy KEe−max the electron can get is
when the nucleus of 55 Mn is created at rest, i.e.
KEe−max = ∆E − me c2
= 3.26 − 0.511
= 2.75 M eV
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
6
CHAPTER 2. RELATIVITY II. HOME WORK SOLUTIONS
2.4
Problem 2.22 (In the text book)
The K ◦ meson is an uncharged member of the particle “zoo” that decays into two charged
pions according to K ◦ −→ π + + π − . The pions have opposite charges, as indicated, and the
same mass, m = 140 M eV /c2 . Suppose that a K ◦ at rest decays into two pions in a bubble
chamber in which a magnetic field of 2.0 T is present (see Figure (2.1)). If the radius of
curvature of the pions is 34.4 cm, find
(a) the momenta and speeds of the pions and
(b) the mass of the K ◦ meson.
Figure 2.1: A sketch of the tracks made by the π + and π − in the decay of K ◦ meson at rest.
The pion motion is perpendicular to B. (B is directed out of the page.)
Solution
(a) The momentum of each pion as it moved in the magnetic field is:
pπ = qπ BRπ
where qπ is the charge of the pion qπ = 1.6 × 10−19 C, B is the magnetic field and Rπ is
the radius of the pion’s path. using the numbers we get:
pπ = 1.6 × 10−19 × 2.0 × 34.4 × 10−2 = 1.1 × 10−19 kg · m/s
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.4. PROBLEM 2.22 (IN THE TEXT BOOK)
7
we need to change the units from kg · m/s to the more relevant units to nuclear reactions
M eV /c. This is done as follows:
kg · m
kg · c · v
=
s
c
kg · m2
= 3 × 108
c · s2
J
= 3 × 108
c
3 × 108 eV
=
1.6 × 10−19 c
3 × 108 × 10−6 M eV
=
1.6 × 10−19
c
M
eV
= 1.875 × 1021
c
The pion momentum is, then:
pπ = 1.1 × 10−19 × 1.875 × 1021
= 206 M eV /c
The K ◦ meson was at rest before the decay, so conservation of momentum requires that
the total momentum after decay be zero. In other words the pions should have equal
and opposite momenta. Since the two pions have equal masses then they should have
equal and opposite velocities. To find the velocity upi of the pions we use pπ = γmπ uπ
and Eπ = γmπ c2 , where p and E are the momentum and total energy of each pion,
pπ
γmπ uπ
=
Eπ
γmπ c2
uπ
= 2
c
uπ
pπ c
=
c
Eπ
= p
= p
pπ c
p2π c2 + (mπ c2 )2
206
(206)2 + (140)2
= 0.827
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
8
CHAPTER 2. RELATIVITY II. HOME WORK SOLUTIONS
(b) Conservation mass energy requires:
EK = 2Eπ
where EK ◦ is the total relativistic energy of the the K ◦ meson and Eπ is the total
relativistic energy of one of the pions. Since K ◦ decayed while at rest we get:
p
mK ◦ c2 = 2 p2π c2 + (mπ c2 )2
p
= 2 (206)2 + (140)2
= 498 M eV
mK ◦ = 498 M eV /c2
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
2.5. PROBLEM 2.29 (IN THE TEXT BOOK)
2.5
9
Problem 2.29 (In the text book)
An object disintegrates into two fragments. One of the fragments has mass 1.00 M eV /c2
and momentum 1.75 M eV /c in the positive x direction. The other fragment has mass 1.50
M eV /c2 and momentum 2.005 M eV /c in the positive y direction. Find
(a) the mass and
(b) the speed of the original object.
Solution
The total energy of the first fragment is:
E1 =
q
p21x c2 + (m1 c2 )2
p
=
(1.75)2 + (1.00)2
= 2.02 M eV
and for the second fragment we have:
q
p22y c2 + (m2 c2 )2
p
=
(2.005)2 + (1.50)2
= 2.50 M eV
E2 =
(a) Mass-energy conservation requires that the total energy of the object before disintegration Eo = E1 + E2 = 4.52 M eV . The momentum after disintegration p is given
by:
p2 = p21x + p22y
= (1.75)2 + (2.00)2
= 7.06 (M eV /c)2
The original object must also have the is momentum, so:
Physics 205:Modern Physics I, Chapter 2
Fall 2004
Ahmed H. Hussein
10
CHAPTER 2. RELATIVITY II. HOME WORK SOLUTIONS
Eo2 = p2 c2 + (mo c2 )2
p
m o c2 =
E 2 − p2 c2
p o
=
(4.52)2 − 7.06
= 3.66 M eV
mo = 3.66 M eV /c2
(b) for the original object we have:
Eo = γmo c2
1
m o c2
=
γ
Eo
r
1−
Physics 205:Modern Physics I, Chapter 2
u2
m o c2
=
c2
Eo
2
2
u
m o c2
= 1−
c2
Eo
s
2
u
mo c2
=
1−
c
Eo
= 0.587
Fall 2004
Ahmed H. Hussein