Physics Solution Manual for 1100 and 2101

Chapter 30 answers to focus on concepts questions

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Unformatted text preview: ve since is moves along the −x direction (the scattering angle is 180°). Using the relation p = h/λ (Equation 29.6), where h is Planck’s constant and λ is the wavelength of the photon, we can write the expression for the momentum of the electron as pelectron = p + p′ = h λ + h 1 1 = h + λ′ λ λ′ SOLUTION Substituting numerical values into the equation above, we have 1 1 pelectron = (6.626 × 10 –34 J ⋅ s) + = 4.755 × 10 –24 kg ⋅ m/s −9 −9 0.2750 × 10 m 0.2825 × 10 m Chapter 29 Problems 1523 20. REASONING The wavelength λ of the incident X-rays is related to the wavelength λ′ of h the scattered X-rays by λ ′ − λ = (1 − cosθ ) (Equation 29.7), where h = 6.626×10−34 J·s is mc 8 Planck’s constant, c = 2.998×10 m/s is the speed of light in a vacuum, m = 9.109×10−31 kg is the mass of an electron, and θ = 122.0° is the angle at which the X-rays are scattered. The wavelength λ′ of the scattered photon is found from the magnitude p′ of its momentum h (Equation 29.6). via p′ = λ′ SOLUTION Solving λ ′ − λ = h (1 − cosθ ) (Equation 29.7) for λ, we obtain mc λ = λ′ − h (1 − cos θ ) mc (1) h h (Equation 29.6) for λ′ yields λ ′ = , which, on substitution into λ′ p′ Equation (1), gives Solving p′ = λ = λ′ − h h h (1 − cos θ ) = − (1 − cos θ ) mc p′ mc ( ( )( )( ) ) 6.626 × 10 −34 J ⋅ s 1 − cos122.0o 6.626 × 10−34 J ⋅ s = − 1.856 × 10 −24 kg ⋅ m/s 9.109 × 10−31 kg 2.998 × 108 m/s = 3.533 × 10 −10 m = 0.3533 nm 21. SSM WWW REASONING The change in wavelength that occurs during Compton scattering is given by Equation 29.7: λ′ – λ = ( λ ′ – λ )max h (1 – cos θ ) mc or ( λ ′ λ )max = h 2h (1 – cos 180° ) = mc mc is the maximum change in the wavelength, and to calculate it we need a value for the mass m of a nitrogen molecule. This value can be obtained from the mass per mole M of nitrogen ( N 2 ) and Avogadro's number N A , according to m = M / N A (see Section 14.1). SOLUTION Using a value of M = 0.0280 kg/mol , we obtain the following result for the maximum change in the wavelength: 1524 PARTICLES AND WAVES ( λ ′ – λ )max ( ) 2 6.63 × 10 –34 J ⋅ s 2h 2h = = = mc M 0.0280 kg/mol 3.00 × 108 m/s N c 23 –1 A 6.02 × 10 mol ( ) = 9.50 × 10 –17 m 22. REASONING Energy is conserved during the collision. This means that the energy E of the incident photon must equal the kinetic energy KE of the recoil electron plus the energy E ′ the scattered photon: E = KE + E ′ (1) The energy E of a photon is related to its frequency f by E = hf (Equation 29.2), where h is Planck’s constant. The frequency, in turn, is related to the wavelength λ by f = c/λ (Equation 16.1), where c is the speed of light in a vacuum. Substituting f = c/λ into E = hf gives (2) E = hc/λ The wavelength λ ′ of the scattered photon depends on the wavelength λ of the incident photon according to Equation 29.7, so that we have λ′ = λ + h (1 − cos θ ) mc Since the photon is scattered straight backward, θ = 180°, and λ′ = λ + h (1 − cos180° ) = λ + 2h mc mc (3) SOLUTION The kinetic energy of the recoil electron is given by KE = 1 mv 2 2 (Equation 6.2), where m is its mass and v is its speed. Substituting this expression into Equation (1), we have E = 1 mv 2 + E ′ . Solving for the speed v of the electron gives 2 v= 2 ( E − E′) m (4) From Equation (2), we also know that E = hc/λ and E ′ = hc / λ ′ . Substituting this expression into Equation (4), we find that the speed of the electron can be written as Chapter 29 Problems v= Since λ ′ = λ + v= = 1525 2h c 1 1 − m λ λ′ 2h [Equation (3)], the speed of the electron is mc 2h c 1 1 λ − 2h m λ+ mc 2 ( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m/s ) 9.11 × 10−31 kg × 1 1 − −9 −34 2 ( 6.63 × 10 J ⋅ s) 0.45000 × 10 m 0.45000 × 10−9 m + −31 8 ( 9.11× 10 kg ) ( 3.00 × 10 m/s ) = 3.22 × 106 m/s 23. REASONING AND SOLUTION a. We have λ = λ' − (h/mc)(1 − cos 163°) = 0.1819 nm b. For the incident photon E = hf = hc/λ = 1.092 × 10 –15 J c. For the scattered photon E ' = hf ' = hc/λ' = 1.064 × 10 –15 J d. The kinetic energy of the recoil electron is, therefore, KE = E − E ' = 2.8 × 10 –17 J 24. REASONING a. Consider one square meter of the sail’s surface. Each of the N photons that strike this square meter in a one-second interval (∆t = 1.0 s) has an initial momentum p that is h determined by its wavelength λ, according to p = (Equation 29.6), where λ 1526 PARTICLES AND WAVES h = 6.63×10−34 J·s is Planck’s constant. Each photon is fully reflected, so the final momentum of a photon is equal to –p. The magnitude of the change ∆p in a photon’s momentum, then, is equal to ∆p = |p − (−p)| = 2p, and the magnitude ∆P of the net momentum change undergone by all N photons is given by ∆P = N∆p = 2Np. In order to cause this momentum change, the sail exerts an impulse of magnitude F∆t = ∆P (Equation 7.4) on the photons, where F is the magnitude of the force exerted...
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