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Unformatted text preview: A − Z = 239 – 94 = 145 neutrons . 4. REASONING According to Coulomb’s law, the magnitude of the electrostatic force
22
between two protons is F = ke /r (Equation 18.1), where e is the magnitude of the charge on
each proton and r is the distance between them. The electrostatic force given by Coulomb’s
law has the least possible magnitude when the two charges are as widely separated as
possible, so that the distance r between them is as great as possible. In the nucleus, this
means that the two protons in question must be located at opposite ends of the diameter of
the nucleus.
SOLUTION To find the magnitude of the least possible electrostatic force that either of two
protons can exert on the other, we need to know the diameter of the gold nucleus. The
diameter is twice the radius, so that Equation 31.2 indicates that the diameter of the gold
97
nucleus is d = 2(1.2 × 10–15 m) A1/3, where A is the nucleon number. For gold 179 Au it
follows that A = 197. Using Coulomb’s law with a separation between the protons that
equals the diameter d of the nucleus, we find FLeast
possible = ke 2
ke 2
=
d2
2 1.2 × 10 −15 m A 1/ 3 c h
8
c.99 × 10 N ⋅ m / C h1.60 × 10
c
=
2 c.2 × 10
1
mh g
197
b
9 2 2 C h=
2 1.2 N SSM WWW REASONING AND SOLUTION Equation 31.2 gives for the radius of the
48
22 Ti nucleus that
r = (1.2 × 10 6. −19 1/ 3 2 −15 5. 2 –15 m)A = (1.2 × 10
1/3 –15 m)(48) = 4.4 × 10 −15 m
1/3 REASONING Assuming that the nuclei are spherical, we will use Area = 4π r 2 to find the
surface area of a nucleus from its radius r. Under the same assumption, the radius r of a nucleus is given by r = (1.2 ×10−15 m ) A1 3 (Equation 31.2), where A is the number of
nucleons in the nucleus.
SOLUTION The ratio of the surface area Area209 of the largest nucleus (A209 = 209) to the
surface area Area1 of the smallest nucleus (A1 = 1) is 1588 NUCLEAR PHYSICS AND RADIOACTIVITY Area 209
Area1 = 2
4π r209 4π r12 = 2
r209 r12 r = 209 r1 2 (1) Substituting r = (1.2 ×10−15 m ) A1 3 (Equation 31.2) into Equation (1), we obtain
2 Area 209
Area1 2
2
23
23 (1.2 × 10 −15 m ) A 1 3 A2091 3 A209 r209 209 209 =
= = = =
13 1 (1.2 × 10 −15 m ) A 1 3 r1 A1 A1 1 = 35.2 7. REASONING The nucleus is roughly spherical, so its volume is V = 4 π r 3 . The radius r is
3
given by Equation 31.2 as r ≈ (1.2 × 10–15 m)A1/3, where A is the atomic mass number or
nucleon number. c Therefore, the volume is given by V = 4 π 1.2 × 10 −15 m
3 hA
3 and is proportional to A. We can apply this expression to the unknown nucleus and to the nickel
nucleus, knowing that the ratio of the two volumes is 2:1. This ratio provides the solution
we seek.
SOLUTION Applying the expression for the volume to each nucleus gives c
c h
hb g
3 −15
4
mA
V
A
3 π 1. 2 × 10
=
=
=2
3
V Ni 4 π 1.2 × 10 −15 m 60
60
3 or A = 120 The nucleon number is equal to the number of neutrons N plus the number of protons or the
atomic number Z, so A = N + Z. Therefore, the atomic number for the unknown nucleus is
Z = A – N = 120 – 70 = 50
The unknown nucleus, then, is 120
50 X . Reference to the periodic table reveals that the element that has an atomic number of Z = 50 is tin (Sn). Thus, 8. A
Z X= 120 Sn .
50 A
REASONING To identify the unknown nucleus in the form Z X , we need to determine the
atomic mass number A (which is the total number of nucleons in the nucleus) and the atomic
number Z (which is the number of protons in the nucleus). Then we can use the periodic
table to identify the nucleus X. To determine A, we will use the given ratio of the nuclear
radii (the radius of a nucleus is proportional to A 1 / 3 ) and the fact that A = 3 for the tritium
nucleus. Since A is the total number of nucleons in the nucleus, the value determined for the
unknown nucleus is the number N of neutrons plus the number Z of protons. We know that Chapter 31 Problems 1589 N is the same for both species and can determine it from the information in the symbol 3 T
1
for the tritium nucleus. Thus, we will be able to determine the value of Z for the unknown
nucleus.
SOLUTION The radius r of a nucleus is ( ) r = 1.2 × 10−15 m A 1/ 3 (31.2) Applying this expression to both nuclei gives (
( )
) 1/3
1.2 × 10−15 m A 1/3 A X
rX
X
=
= 1/3
1/3
rT
AT
1.2 × 10−15 m A T or rX r
T 3 AX = AT Solving for AX, we find that
3 r 3
A X = AT X = 3 (1.10 ) = 4
r T
Since AX is the number N of neutrons plus the number Z of protons in the unknown nucleus,
it follows that
4= N +Z The value of N is the same as it is for the tritium nucleus 3 T , for which 3 = N + 1, so that
1
N = 2. Thus, for the unknown nucleus, we have 4 = N +Z = 2+Z or Z = 4−2 = 2 From the periodic table, we identify the nucleus for which Z = 2 as the helium (He) nucleus:
4
2 He 9. SSM WWW REASONING According to Equation 31.2, the radius of a nucleus in c h meters is r = 1.2 × 10 –15 m A 1/ 3 , where A is the nucleon number. If we treat the neutron
star as a uniform sphere, its density (Equation 11.1) can be written as ρ=
Solving for the radius r, we obtain, M
=
V 4
3 M
πr3 1590 NUCLEAR PHYSICS AND...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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