2
Problem
:
For lead (Pb) find (a) the net electrical charge of
nucleus, (b) the number of neutrons, (c)the number of nucleons,
(d) the approximate radius of nucleus, and (e) the nuclear
density.
a. The net electrical charge of the nucleus is equal to the total number of
protons multiplied by the charge on a single proton:
For lead, we have A = 208 and Z = 82.
C
C
q
net
19
19
10
31
.
1
)
10
6
.
1
(
82
−
−
×
+
=
×
+
=
b.
The number of neutrons is N = A – Z = 208 – 82 = 126
c.
By inspection, the number of nucleons is A = 208
d. The approximate radius of the nucleus
rA
=×
(.
/
12 10
10
13
–15
–15
1/3
–15
m)
(1.2
10
m)(208)
7.1
m
Problem (con’d)
:
For lead (Pb) find (a) the net
electrical charge of nucleus, (b) the number of
neutrons, (c)the number of nucleons, (d) the
approximate radius of nucleus, and (e) the nuclear
density.
For lead, we have A = 208 and Z = 82.
e. The nuclear density is the mass per unit volume of the nucleus.
The
total mass of the nucleus can be found by multiplying the mass
of a single
nucleon by the total number
A
of nucleons in the nucleus.
Treating the
nucleus as a sphere of radius
r
, the nuclear density is
ρ
π
==
=
×
=
×
m
V
mA
r
A
m
total
nucleon
nucleon
–15
nucleon
–15
3
m)
m)
4
3
3
4
3
3
4
3
/
=
×
×
167 10
10
4
3
.
–27
–15
3
17
3
kg
m)
2.3
kg / m
The atomic mass
unit
±
The
atomic mass unit
,
u
,
is one-twelfth of the mass of
12
C
atom of carbon.
±
The
atomic mass unit
,
u
,
is about the mass of one proton
or neutron.The atomic masses of the elements in
u
are
average masses, taking into account the different isotopes
that exist.
±
The energy equivalent of one atomic mass unit:
u =
1.4924 x 10
-10
J = 931.5 MeV;
±
One gram is about 600,000,000,000,000,000,000,000 u (that's 600
sextillion, or a 6 x 10
23
).
±
A pound is just 300 septillion u or 3 x 10
26
or
300,000,000,000,000,000,000,000,000.
±
Now you can see why we use u's instead of grams!!