Physics Solution Manual for 1100 and 2101

The de broglie wavelength of one electron in the beam

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Unformatted text preview: he ice phase to the liquid phase is N ′ 6.7 × 10 25 molecules = = 32 molecules / photon N 2.1 × 10 24 photons 15. SSM REASONING The angle θ through which the X-rays are scattered is related to the difference between the wavelength λ ′ of the scattered X-rays and the wavelength λ of the incident X-rays by Equation 29.7 as 1520 PARTICLES AND WAVES λ′ − λ = h (1 − cos θ ) mc where h is Planck’s constant, m is the mass of the electron, and c is the speed of light in a vacuum. We can use this relation directly to find the angle, since all the other variables are known. SOLUTION Solving Equation 29.7 for the angle θ, we obtain cos θ = 1 − mc (λ′ − λ ) h ( 9.11×10−31 kg ) ( 3.00 ×108 m/s ) ( 0.2703 ×10−9 m − 0.2685 ×10−9 m ) = 0.26 = 1− 6.63 ×10−34 J ⋅ s θ = cos −1 ( 0.26 ) = 75° 16. REASONING The momentum of the photon is related to its wavelength λ and Planck’s constant h. The momentum (nonrelativistic) of the ball depends on its mass m and speed v. We can set the two momenta equal and solve directly for the speed. SOLUTION The momentum pphoton of the photon and the momentum pball of the ball are pphoton = h λ (29.6) and pball = mv (7.2) Since pphoton = pball, we have h 1 24 4λ 3 Momentum of photon = mv 1 24 43 Momentum of ball or v= h 6.63 ×10−34 J ⋅ s = = 4.2 ×10−25 m/s −9 −3 λ m 720 ×10 m 2.2 × 10 kg ( )( ) 17. REASONING The frequency f of a photon is related to its energy E by f = E/h (Equation 29.2), where h is Planck’s constant. As discussed in Section 29.4, the energy E is related to the magnitude p of the photon’s momentum by E = pc, where c is the speed of light in a vacuum. By combining these two relations, we see that the frequency can be expressed in terms of p as f = pc/h. SOLUTION a. Substituting values for p, c, and h, into the relation f = pc/h gives Chapter 29 Problems f= 1521 pc ( 2.3 × 10−29 kg ⋅ m/s ) ( 3.00 ×108 m/s ) = = 1.0 × 1013 Hz −34 h 6.63 ×10 J ⋅s b. An inspection of Figure 24.9 shows that this frequency lies in the infrared region of the electromagnetic spectrum. 18. REASONING Before the scattering, the electron is at rest and has no momentum. Thus, the total initial momentum consists only of the photon’s momentum, which points along the +x axis. The total initial momentum has no y component. Since the total momentum is conserved, the total momentum after the scattering must be the same as it was before and, therefore, has no y component. The total momentum after the scattering is the sum of the momentum of the scattered photon and that of the scattered electron, and it only has an x component. But the scattered photon is moving along the –y axis, so its momentum has no x component. Therefore, the momentum of the electron must have an x component. The total momentum after the scattering is the sum of the momentum of the scattered photon and that of the scattered electron, and it has no y component. But the scattered photon is moving along the –y axis, so its momentum points along the –y axis. Therefore, this contribution to the total final momentum must be cancelled by part of the momentum of the scattered electron, which must have a component along the +y axis. SOLUTION Since the total momentum is conserved and since the scattered photon has no momentum in the x direction, the momentum of the scattered electron must have an x component that equals the momentum of the incident photon. According to Equation 29.6, the magnitude p of the momentum of the incident photon is p = h/λ, where h is Planck’s constant and λ is the wavelength. Therefore, the momentum of the scattered electron has a component in the +x direction that is px = h λ = 6.63 × 10 −34 J ⋅ s = 7 .37 × 10 −23 kg ⋅ m / s −12 9.00 × 10 m The momentum of the scattered electron has a component along the +y direction. This component cancels the momentum of the scattered photon that points along the –y direction. To find the momentum of the scattered photon, we first need to determine its wavelength λ′, which we can do using Equation 29.7: 1522 PARTICLES AND WAVES h λ′ = λ + 1 − cos 90.0° 43 mc 1 24 =0 = 9.00 × 10−12 m + 6.63 × 10−34 J ⋅ s (9.11 × 10 −31 )( kg 3.00 × 10 m/s 8 ) = 1.14 × 10−11 m Again using Equation 29.6, we find that the momentum of the scattered electron has a component in the +y direction that is py = h λ = 6.63 × 10 −34 J ⋅ s = 5.82 × 10 −23 kg ⋅ m / s −11 1.14 × 10 m 19. REASONING There are no external forces that act on the system, so the conservation of linear momentum applies. Since the photon is scattered at θ = 180 ° , the collision is "head-on," and all motion occurs along the horizontal direction, which we take as the x axis. The incident photon is assumed to be moving along the +x axis. For an initially stationary electron, the conservation of linear momentum states that: p = − p′ + pelectron 1 23 1 24 43 1 24 43 Momentum of incident photon Momentum of scattered photon Momentum of recoil electron where the momentum of the scattered photon is negati...
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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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