1.1
DOING PHYSICS
EXERCISES
Section 1.2 Measurements and Units
10.
SI prefixes are explained in Table 1.1.
(a)
10
10 10
10
33
6
9
MW
W
W.
==
()
(b)
and
(c)
10
10
1
96
Wk
W
G
W
.
11.
INTERPRET
We interpret this as a problem involving the comparison of the diameter of two objects (a hydrogen
and a proton) expressed in different units.
DEVELOP
Before any comparison can be made, the quantity of interest must first be expressed in the same units.
From Table 1.1, we see that a nanometer is10
9
−
m, and a femtometer (fm) is10
15
−
m.
EVALUATE
Using the conversion factors, the diameter of a hydrogen atom is
d
H
nm
m,
−
01
10
10
.
and the
diameter of a proton is
d
p
fm
−
11
0
15
Therefore, the ratio of the diameters of a hydrogen atom and a proton
(its nucleus) is
d
d
H
p
m
m
−
−
10
10
10
10
15
5
ASSESS
The hydrogen atom is about 100,000 times larger than its nucleus.
12.
The current definition of the meter is such that the speed of light in vacuum is exactly 299,792,458 m/s. The
distance traveled is the speed multiplied by the travel time,
0
9
ns
s
=
−
in this case. Thus
d
=×
=
−
(,,
/
)
299 792 458
10
9
m s
s
0.299792458 m, or approximately 30 cm.
13.
INTERPRET
We interpret this as a problem involving expressing the period of cesium radiation in different units.
DEVELOP
By definition, 1
9 192 631 770
s
=
,,,
periods of a cesium atomic clock. In addition, we know that
0
9
ns
s.
=
−
EVALUATE
One period of cesium radiation is
1
9 192 631 770
1 087827757 10
10
s
s
0.1087
.
=
−
827757 ns
ASSESS
Since one nanosecond corresponds to about 9 periods of the cesium radiation, each period is about
1
9
of
a nanosecond. Note that there exists an alternative definition based on the frequency of the cesium-133 hyperfine
transition, which is the reciprocal of the period.
14.
The prefix
eka
equals 10
18
, so 14 Eg
14
10
(10
kg)
1.4
10
kg.
18
3
16
= ×
−
15.
INTERPRET
We interpret this as a problem involving expressing 1-cm line as multiples of the diameter of
hydrogen atoms.
DEVELOP
We first express the quantities of interest (diameter of a hydrogen atom and1-cm line) in the same
units. Since a nanometer is 10
9
−
m (Table 1.1), we see that
d
H
nm
m.
−
10
.
In addition,
0
2
cm
=
−
EVALUATE
The desired number of atoms is the length of the line divided by the diameter of one atom:
N
−
−
10
10
10
2
10
8
m
m
ASSESS
If 1 cm corresponds to10
8
hydrogen atoms, then each atom would correspond to 10
cm
8
−−
=
10
10
m
=
0.1 nm.
1