Physics Solution Manual for 1100 and 2101

Thus for the unknown nucleus we have 4 n z 2z or z

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Unformatted text preview: age does not change, the cutoff wavelength remains the same, independent of the target material. The wavelength of the Kα photon is given by Equation 30.14, with ni = 2 and nf = 1. The wavelength λ is proportional to 1/(Z – 1)2, so the wavelength decreases for larger values of Z. Since silver has a larger value of Z than molybdenum, the wavelength of the Kα photon decreases when silver replaces molybdenum. SOLUTION a. The cutoff wavelength λ0, which is the same for either target material, is ( )( )( ) 6.63 × 10−34 J ⋅ s 3.00 × 108 m/s hc λ0 = = = 3.55 × 10−11 m −19 3 eV 1.60 × 10 C 35.0 × 10 V ( ) (30.17) b. The Kα photon is emitted when an electron drops from the ni = 2 to the nf = 1 level. The wavelengths emitted by molybdenum and silver are: ( ) 1 2 1 = 1.097 × 107 m −1 ( 42 − 1) 2 − 2 = 1.38 × 1010 m −1 λ 2 1 1 Molybdenum (30.14) λ = 7.23 × 10−11 m Silver ( ) 1 2 1 = 1.097 × 107 m −1 ( 47 − 1) 2 − 2 = 1.74 × 1010 m −1 λ 2 1 1 λ = 5.74 × 10−11 m (30.14) 1570 THE NATURE OF THE ATOM ______________________________________________________________________________ 41. REASONING AND SOLUTION a. The energy difference between the two Kβ energy levels is E = hf = hc/λ, according to Equation 29.2 and Equation 16.1: E= hc λ = ( 6.63 ×10−34 J ⋅ s ) ( 3.00 ×108 m/s ) 1.84 × 10−11 m = 1.08 ×10−14 J b. Converting to electron volts, we have (1.08 ×10−14 J ) 1 eV = 6.75 ×104 eV −19 1.60 ×10 J ______________________________________________________________________________ 42. REASONING We know from our study of waves (see Section 16.2) that the wavelength of a wave is inversely proportional to its frequency. As discussed in Section 29.3, the frequency of a wave, such as an X–ray photon, is directly proportional to its energy. The energy needed to create the wave comes from the kinetic energy of the electron impinging on the metal target. We can determine the electron’s kinetic energy from its speed, since the two are related. Thus, we will be able to evaluate the wavelength of the X-ray photon from a knowledge of the electron’s speed. SOLUTION As discussed in Section 16.2, the wavelength λ of a wave is related to its frequency f and speed v by λ = v / f (Equation 16.1). In a vacuum, an electromagnetic wave travels at the speed c of light. Substituting v = c into λ = v / f gives λ = c / f . An X-ray photon is an electromagnetic wave that is a discrete packet of energy. The photon’s frequency f is related to its energy E by f = E / h (Equation 29.2), where h is Planck’s constant. Substituting this expression for f into λ = c / f gives λ= c c ch = = f E E h (1) The energy needed to produce an X-ray photon comes from the kinetic energy of an electron striking the target. If the speed of the electron is much less than the speed of light in a vacuum, its kinetic energy KE given by Equation 6.2 as KE = 1 mv 2 , where m and v are the 2 mass and speed of the electron. We are given that the electron decelerates to one-quarter of its original speed, so that its loss of kinetic energy is Loss of kinetic energy = 1 mv 2 2 −1m 2 ( 1 v) 4 2 = 15 mv 2 32 Chapter 30 Problems 1571 The kinetic energy lost by the decelerating electron goes into creating the X-ray photon, so that E = 15 mv 2 . Substituting this expression for E into Equation (1) gives 32 λ= ch = E ch 15 mv 2 32 = ( 3.00 ×108 m/s ) ( 6.63 ×10−34 J ⋅ s ) 2 15 ( 9.11× 10 −31 kg ) ( 6.00 × 107 m/s ) 32 = 1.29 ×10−10 m ______________________________________________________________________________ 43. SSM REASONING In the spectrum of X-rays produced by the tube, the cutoff wavelength λ0 and the voltage V of the tube are related according to Equation 30.17, V = hc / ( eλ0 ) . Since the voltage is increased from zero until the Kα X-ray just appears in the spectrum, it follows that λ0 = λα and V = hc / ( eλα ) . Using Equation 30.14 for 1 / λα , we find that V= 2 hc h cR ( Z –1) 1 = 2 1 eλα e 1 22 In this expression we have replaced Z with Z − 1, in order to account for shielding, as explained in Example 11 in the text. SOLUTION The desired voltage is, then, ( 6.63 × 10–34 J ⋅s ) (3.00 × 108 m/s ) (1.097 × 107 m –1 ) ( 47–1)2 1 – 1 = V= 2 1 22 (1.60 × 10–19 C ) 21 600 V ______________________________________________________________________________ 44. REASONING Using Equation 30.13 for the total energy E1 of an electron in the K-shell (n = 1) and using Z − 1 instead of Z to account for shielding (see Example 10), we have E1 = − (13.6 eV ) ( Z − 1)2 12 When striking the metal target, an incoming electron must have at least enough energy to raise the K-shell electron from this low energy level up to the 0-eV level that corresponds to a very large distance from the nucleus. Only then will the incoming electron knock the K-shell electron entirely out of a target atom. Thus, the value for the minimum energy Emin that an incoming electron must have is Emin = (13.6 eV...
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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.

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