which is equivalent to the result shown in the problem statement. Since the desired
numerical answer uses eV units, we multiply numerator and denominator of our result by
c
2
and make use of the
mc
2
value for an electron in Table 373 as well as the value
1240eV nm
hc
=⋅
:
NE
mc
hc
nn
n
F
()
(
(
(.
)
//
/
=
F
H
G
I
K
J
=
×
⋅
F
H
G
I
K
J
−−
4
3
4 511 10
1240
34
1
1
2
2
2
3
13
3
2
3
2
1
ππ
eV)
eV nm)
nm
eV
2
which is equivalent to the value indicated in the problem statement.
(b) Since there are 10
27
cubic nanometers in a cubic meter, then the result of Problem 41
5 may be written as
n
=×
=
849 10
849
28
3
3
..
.
mn
m
The cube root of this is
n
1/3
≈
4.4/nm. Hence, the expression in part (a) leads to
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 Spring '08
 Any
 Physics

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