Physics Solution Manual for 1100 and 2101

Since l must be less than n this subshell

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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 ______________________________________________________________________________ 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 cross-sectional area of the nucleus to the cross-sectional 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 kinetic-energy 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 ______________________________________________________________________________ r= 6. REASONING The work WAB done o...
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