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PHYS 133 WEEK 3
CHAPTER 28
28.3.
Solve:
Using Equation 28.3 and Table 28.1, the electron current is
i
=
nAv
d
=
5.9
!
10
28
m
"
3
(
)
#
0.5
!
10
"
3
m
(
)
2
5.0
!
10
"
5
m / s
(
) =
2.3
!
10
18
s
"
1
The time for 1 mole of electrons to pass through a cross section of the wire is
t
=
N
A
!
1 mole
i
=
6.02
!
10
23
2.3
!
10
18
s
"
1
=
2.62
!
10
5
s
#
3.03 days
Assess:
The drift speed is small, and Avogadro’s number is large. A time of the order of 3 days is reasonable.
28.7.
Solve:
(a)
Each gold atom has one conduction electron. Using Avogadro’s number and
n
as the number of moles, the
number of atoms is
N
=
nN
A
=
m
M
A
N
A
=
!
V
M
A
N
A
=
!"
r
2
L
( )
M
A
N
A
The density of gold is
ρ
=
19,300 kg/m
3
, the atomic mass is
M
A
=
197 g mol
−
1
,
r
=
0.5
×
10
−
3
m,
L
=
0.1 m, and
N
A
=
6.02
×
10
23
mol
−
1
. Substituting these values, we get
N
=
4.63
×
10
21
electrons.
(b)
If all the electrons are transferred a charge of (4.63
×
10
21
)(
−
1.60
×
10
−
19
C)
=
−
740.8 C will be delivered. To deliver a
charge of
−
32 nC, however, the electrons within a length
l
have to be delivered. Thus,
l
=
!
32
"
10
!
9
C
!
740.8 C
10 cm
( ) =
4.32
"
10
!
10
cm
=
4.32
"
10
!
12
m
28.11.
Model:
A battery is a charge escalator.
Solve:
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 Spring '04
 Saunders
 Current

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