Physics Solution Manual for 1100 and 2101

# Chapter 30 answers to focus on concepts questions

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online