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Unformatted text preview: RADIOACTIVITY r= 3 M
4
3 πρ This expression can be used to find the radius of a neutron star of mass M and density ρ .
SOLUTION As discussed in Conceptual Example 1, nuclear densities have the same
approximate value in all atoms. If we consider a uniform spherical nucleus, then the density
of nuclear matter is approximately given by ρ= = M
A × (mass of a nucleon)
≈
=
3
4
V
3πr A × (mass of a nucleon)
4
3 π (1.2 × 10 –15 m ) A 1/ 3 3 1.67 × 10 –27 kg
= 2 .3 × 10 17 kg / m 3
–15
3
4
m)
3 π ( 1.2 × 10 The mass of the sun is 1.99 × 1030 kg (see inside of the front cover of the text). Substituting
values into the expression for r determined above, we find
r= 3 ( 0.40 )( 1.99 × 10 30 kg)
=
17
4
kg / m 3 )
3 π ( 2 .3 × 10 9.4 × 10 3 m 10. REASONING
a. The neutrons packed into the tennis ball are like a giant spherical nucleus containing N
neutrons. Because no protons are present, the atomic number Z of the sphere is zero, and the
atomic mass number A is equal to N, as we see from A = Z + N = 0 + N = N (Equation 31.1).
Because the neutrons are packed into a spherical shape as densely as the nucleons in a
nucleus, the mass number A of the sphere is related to its radius by r = (1.2 ×10−15 m ) A1 3
(Equation 31.2). Substituting N = A into Equation 31.2 gives r = (1.2 ×10−15 m ) N 1 3 (1) b. According to Newton’s 2nd law, the magnitude a of the object’s acceleration is equal to
the magnitude F of the gravitational force exerted on it divided by its mass m: a = F/m. The
Mm
magnitude of that gravitational force is given by F = G 2 (Equation 4.3), where
R
G = 6.67×10−11 N·m2/kg2 is the universal gravitational constant, R = 2.0 m is the distance
between the object and the center of the tennis ball, and M is the mass of the tennis ball. The
mass of the tennis ball is the mass mn = 1.67×10−27 kg of a single neutron multiplied by the
number N of neutrons found in part (a), so we have that Chapter 31 Problems 1591 M = Nmn (2) SOLUTION
a. Cubing both sides of Equation (1) and solving for N, we obtain r 3 3 3
= (1.2 ×10−15 m) N b. Substituting F = G or 3 r 0.032 m 40
N = = = 1.9 ×10
−15
−15 1.2 ×10 m 1.2 ×10 m Mm
F
into a = , we find that
2
m
R a= F GM m GM
=2
=
m R2 m
R (3) Substituting Equation (2) into Equation (3) yields a= GM GNmn ( 6.67 ×10−11 N ⋅ m2 /kg 2 ) (1.9 ×1040 ) (1.67 ×10−27 kg )
=
=
= 530 m/s 2
R2
R2
( 2.0 m )2 11. SSM REASONING To obtain the binding energy, we will calculate the mass defect and
then use the fact that 1 u is equivalent to 931.5 MeV. The atomic mass given for 7 Li
3
includes the 3 electrons in the neutral atom. Therefore, when computing the mass defect, we
must account for these electrons. We do so by using the atomic mass of 1.007 825 u for the
hydrogen atom 1 H , which also includes the single electron, instead of the atomic mass of a
1
proton.
SOLUTION Noting that the number of neutrons is 7 – 3 = 4, we obtain the mass defect ∆m
as follows: ∆ m = 3 (1.007 825 u ) + 4 (1.008 665 u ) −
14 244
4
3
14 244
4
3
3 hydrogen atoms
(protons plus electrons) 7.016 003 u
14243 = 4.2132 ×10−2 u Intact lithium atom
(including 3 electrons) 4 neutrons Since 1 u is equivalent to 931.5 MeV, the binding energy is ( ) 931.5 MeV Binding energy = 4.2132 ×10−2 u = 39.25 MeV
1u 1592 NUCLEAR PHYSICS AND RADIOACTIVITY 12. REASONING The mass defect is the total mass of the stationary separated nucleons
(protons and neutrons) minus the mass of the intact nucleus. The given atomic masses are
for the electrical neutral atoms and, therefore, include the mass of the electrons. This will
cause no problem, provided that we use the atomic mass of the hydrogen atom (including its
electron) when determining the mass of the separated nucleons. Referring to Table 31.1 we
find that the mass of a neutron is 1.008 665 u and the mass of a hydrogen atom is
1.007 825 u.
SOLUTION
a. The helium 3
2 He nucleus contains 2 protons and 3 − 2 = 1 neutron. Thus, the mass defect ∆m is
∆ m = 2 (1.007 825 u ) + 1(1.008 665 u ) − 3.016 030 u = 0.008 285 u
144444 2444444 14243
4
3
Separated nucleons: 2 protons and 1 neutron Intact nucleus b. The tritium 3 T nucleus contains 1 proton and 3 − 1 = 2 neutrons. Thus, the mass defect
1 ∆m is
∆ m = 1(1.007 825 u ) + 2 (1.008 665 u ) − 3.016 050 u = 0.009 105 u
144444 2444444 14243
4
3
Separated nucleons: 1 proton and 2 neutrons Intact nucleus 3
c. The mass defect for tritium 3 T is greater than that for helium 2 He . The mass defect is
1 related to the binding energy as follows: Binding energy = ( ∆ m ) c 2 (31.3) The binding energy is the energy that must be supplied to the intact nucleus in order to
separate it into its constituent nucleons. Thus,
3
more energy must be supplied to tritium 3 T than to helium 2 He .
1 13. REASONING AND SOLUTION For 202
80 Hg the mass of the separated nucleons is m = 80(1.007 825 u) + 122(1.008 665 u) = 203.683 130 u
The mass defect is then ∆m = 203.683 130 u – 201.970 617 u = 1.712 513 u. Chapter 31 Problems b This corresponds to a total binding...
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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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