P27_009.PDF
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66 KB
)
Pobierz
Chapter 27 - 27.9
9. (a) The charge that strikes the surface in time ∆
t
is given by ∆
q
=
i
∆
t
,where
i
is the current. Since
each particle carries charge 2
e
, the number of particles that strike the surface is
N
=
∆
q
2
e
=
i
∆
t
=
(0
.
25
×
10
−
6
A)(3
.
0s)
=2
.
3
×
10
12
.
2
e
2(1
.
6
×
10
−
19
C)
(b) Now let
N
be the number of particles in a length
L
of the beam. They will all pass through the
beam cross section at one end in time
t
=
L/v
,where
v
is the particle speed. The current is the
charge that moves through the cross section per unit time. That is,
i
=2
eN/t
=2
eNv/L
.Thus
N
=
iL/
2
ev
. To find the particle speed, we note the kinetic energy of a particle is
10
−
12
J
.
Since
K
=
2
mv
2
, then the speed is
v
=
2
K/m
. The mass of an alpha particle is (very nearly) 4
times the mass of a proton, or
m
=4(1
.
67
K
=20MeV=(20
×
10
6
eV)(1
.
60
×
10
−
19
J
/
eV) = 3
.
2
×
×
10
−
27
kg) = 6
.
68
×
10
−
27
kg, so
v
=
2(3
.
2
×
10
−
27
kg
=3
.
1
10
−
12
J)
×
10
7
m
/
s
6
.
68
×
and
iL
2
ev
=
(0
.
25
×
10
−
6
)(20
×
10
−
2
m)
N
=
10
7
m
/
s)
=5
.
0
×
10
3
.
2(1
.
60
×
10
−
19
C)(3
.
1
×
10
−
12
J. We note, too, that the initial potential energy is
U
i
=
qV
=2
eV
,and
the final potential energy is zero. Here
V
is the electric potential through which the particles are
accelerated. Consequently,
×
K
f
=
U
i
=2
eV
=
⇒
V
=
K
f
2
e
=
3
.
2
×
10
−
12
J
10
−
19
C)
=10
×
10
6
V
.
2(1
.
60
×
(c) We use conservation of energy, where the initial kinetic energy is zero and the final kinetic energy
is 20MeV = 3
.
2
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