This preview shows page 1. Sign up to view the full content.
580
Nuclear Structure
P44.34
(a)
A gamma ray has zero charge and it contains no protons or neutrons. So for a gamma ray
Z
=
0 and
A
=
0. Keeping the total values of
Z
and
A
for the system conserved then requires
Z
=
28 and
A
=
65 for
X
. With this atomic number it must be nickel, and the nucleus must be
in an exited state, so it is
28
65
Ni
*
.
(b)
α
=
2
4
He has
Z
=
2
and
A
=
4
so for
X
we require
Z
=−=
84 2
82
for Pb
and
A
=−
=
215 4
211,
X
=
82
211
Pb.
(c)
A positron e
e
+
=
1
0
has charge the same as a nucleus with
Z
=
1. A neutrino
0
0
ν
has no charge.
Neither contains any protons or neutrons. So
X
must have by conservation
Z
=+
=
26 1
27. It
is Co. And
A
=+=
55 0
55. It is
27
55
Co.
Similar reasoning about balancing the sums of
Z
and
A
across the reaction reveals:
(d)
−
1
0
e
(e)
1
1
H (or p). Note that this process is a nuclear reaction, rather than radioactive decay. We can
solve it from the same principles, which are fundamentally conservation of charge and
conservation of baryon number.
P44.35
N
C
=
F
H
G
I
K
J
×
00210
602 10
23
.
.
g
12.0 g mol
molecules mol
ej
N
C
=×
105 10
21
.
carbon atoms
of which 1 in 7 70 10
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '11
 Staff
 Physics, Charge, Neutron

Click to edit the document details