This preview shows page 1. Sign up to view the full content.
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
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics

Click to edit the document details