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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 Xray 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
Xray 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 Xray 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 onequarter 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 Xray 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 Xrays 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 Xray 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 Kshell
(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 Kshell electron from this low energy level up to the 0eV level that corresponds to
a very large distance from the nucleus. Only then will the incoming electron knock the
Kshell 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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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