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Unformatted text preview: n is greater when the positive nuclear charge is
greater.
15. 2.3 × 10−11 m
16. (b) According to Equation 30.17, the cutoff wavelength is inversely proportional to the
voltage applied across the tube. Therefore, when the voltage increases by a factor of two, the
cutoff wavelength decreases by a factor of two.
17. (d) This statement correctly describes a population inversion (see Section 30.8).
18. (d) The population inversions used in a lasers involve a higher (not lower) energy state that
is metastable (see Section 30.8). 1546 THE NATURE OF THE ATOM CHAPTER 30 THE NATURE OF THE ATOM
PROBLEMS
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1. SSM REASONING Assuming that the hydrogen atom is a sphere of radius ratom, its
volume Vatom is given by
volume Vnucleus is given by 4
3
π ratom . Similarly,
3
4
3
π rnucleus .
3 if the radius of the nucleus is rnucleus, the SOLUTION
a. According to the given data, the nuclear dimensions are much smaller than the orbital
radius of the electron; therefore, we can treat the nucleus as a point about which the electron
–11
orbits. The electron is normally at a distance of about 5.3 × 10 m from the nucleus, so we
can treat the atom as a sphere of radius ratom = 5.3 ×10−11 m . The volume of the atom is ( 3
Vatom = π ratom = π 5.3 × 10−11 m
4
3 4
3 ) 3 = 6.2 ×10−31 m3 b. Similarly, since the nucleus has a radius of approximately rnucleus = 1×10−15 m , its
volume is ( 3
Vnucleus = π rnucleus = π 1× 10−15 m
4
3 4
3 ) 3 = 4 ×10−45 m3 c. The percentage of the atomic volume occupied by the nucleus is
3 4 π r3
3 nucleus
4 π r3
3 atom 1×10−15 m −13
×100% =
× 100% = % 5.3 × 10−11 m × 100% = 7 ×10 Vatom ______________________________________________________________________________
Vnucleus 2. REASONING The area A of a circle of radius R is given by A = π R 2 , so that the crosssectional areas of the target atom and nucleus are
2
Anucleus = π Rnucleus and 2
Aatom = π Ratom (1) We will use Equations (1) and the given ratio Anucleus Aatom = 2.6 ×10−7 to determine the
radius Rnucleus of the nucleus. SOLUTION From Equations (1), the ratio of the crosssectional area of the nucleus to the
crosssectional area of the atom is Chapter 30 Problems Anucleus
Aatom = 2
π Rnucleus = 2
π Ratom 2
Rnucleus 1547 (2) 2
Ratom Solving Equation (2) for Rnucleus, we obtain
A
2
2
Rnucleus = Ratom nucleus
A atom
Rnucleus = Ratom Anucleus
Aatom or ( = 1.4 × 10 −11 m ) 2.6 × 10 −7 = 7.1 × 10 −15 m ______________________________________________________________________________
3. REASONING In the absence of relativistic effects, the kinetic energy is KE = 1 mv 2 , where
2
m is the mass of the α particle and v is its speed. We can relate the kinetic energy to the
de Broglie wavelength λ by recognizing two things. The first is that λ = h/p, according to
Equation 29.8, where h is Planck’s constant and p is the magnitude of the α particle’s
momentum. The second is that the magnitude of the momentum is p = mv. With this
substitution for p, the de Broglie wavelength becomes λ= h
h
=
p mv or mv = h λ Substituting the expression for mv into the kineticenergy expression gives m 2v 2 ( h / λ )
h2
=
=
=
2m
2m
2mλ 2
2 KE = 1 mv 2
2 (1) SOLUTION Using Equation (1), we find that ( ) 2 6.63 ×10−34 J ⋅ s
h2
KE =
=
2mλ 2 2 6.64 × 10−27 kg 1.4 ×10−14 m ( 4. )( ) 2 = 1.7 ×10−13 J REASONING According to the discussion in Conceptual Example 1, the radius of the
electron orbit is about 1 × 105 times as great as the radius of the nucleus. In the scale
5
model, the radius of the electron’s orbit must also be 1 × 10 times as great as the radius of
ball. 1548 THE NATURE OF THE ATOM SOLUTION Since the ball that represents the nucleus has a radius of 3.2 cm, the distance
between the nucleus and the nearest electron in the model must be = 2 mi 1.61 × 10 cm ______________________________________________________________________________ ( 3.2 cm ) (1 × 105 ) = ( 3.2 × 105 cm ) 5. 1 mi 5 SSM REASONING The distance of closest approach can be obtained by setting the
kinetic energy KE of the α particle equal to the electric potential energy EPE of the α
particle. According to Equation 19.3, we have that EPE = (2e)V, where 2e is the charge on
the α particle and V is the electric potential created by a gold nucleus. According to
Equation 19.6, the electric potential of the gold nucleus (charge = Ze =79e) is V = k(79e)/r,
where r is the distance between the α particle and the gold nucleus. Therefore, we have that EPE = ( 2e ) V = ( 2e ) k ( 79e )
r (1) In this expression, we note that k = 8.99 × 109 N ⋅ m 2 /C2 and e = 1.602 × 10−19 C .
SOLUTION Solving Equation (1) for the distance r we obtain ( 2e ) k ( 79e ) = (8.99 ×109 N ⋅ m 2 /C2 )2(79)(1.602 × 10−19 C) 2
= 7.3 ×10−14 m
EPE
5.0 × 10−13 J
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r= 6. REASONING The work WAB done o...
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