Physics Solution Manual for 1100 and 2101

# 63 1034 j s h2 ke 2m 2 2 664 1027 kg 14 1014 m 4 2

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Unformatted text preview: on the photons per square meter of the sail. By Newton’s Third Law, that force magnitude is equal to the magnitude of the force exerted by the N photons on one square meter of the sail. From Equation 7.4, then, we have that F ∆t = ∆P = 2 Np (1) Lastly, the magnitude ΣF = F of the net force necessary for the sail to attain the desired acceleration of a = 9.8×10−6 m/s2 is given by Newton’s Second Law, F = ma (Equation 4.1), where m = 3.0×10−3 kg is the mass of one square meter of the sail. As instructed, we have ignored all other forces acting on the sail. b. The intensity S of the laser beam depends on the total energy delivered to the sail by T otal energy (Equation 24.4), where A is the area of the sail and ∆t is the time interval. S= ∆t A The total energy is equal to the number N of photons that strike the area in one second times the energy E of a single photon. Therefore, the intensity of the laser beam is S= T otal energy NE = ∆t A ∆t A (2) The energy E of a single photon is given by E = hf (Equation 29.2), where f = c λ (Equation 16.1) is the photon’s frequency and c = 3.00×10 m/s is the speed of light in a vacuum. 8 SOLUTION a. Solving Equation (1) for N, we obtain N= Substituting p = h λ F ∆t 2p (Equation 29.6) and F = ma (Equation 4.1) into Equation (3) yields (3) Chapter 29 Problems N= 1527 F ∆t ma∆t ma∆tλ = = 2p 2h h 2 λ (3.0 ×10−3 kg ) ( 9.8 ×10−6 m/s2 ) (1.00 s ) ( 225 ×10−9 m ) = 5.0 ×1018 = 2 ( 6.63 ×10−34 J ⋅ s ) b. Substituting f = c (Equation 16.1) into E = hf (Equation 29.2) gives E = hf = λ Substituting this result into Equation (2), we obtain S= NE Nhc = ∆ t A ∆ t Aλ hc λ . (4) Equation (4) applies to the intensity reaching an area A = 1.0 m2 of the sail in a time ∆t = 1.0 s. Therefore, in order for the sail to accelerate at the desired rate, the intensity of the laser must be Nhc ( 5.0 ×1018 ) ( 6.63 ×10−34 J ⋅ s ) ( 3.00 ×108 m/s ) S= = = 4.4 W/m2 2 )( −9 ∆t Aλ (1.00 s ) (1.00 m 225 ×10 m ) 25. SSM REASONING AND SOLUTION The de Broglie wavelength λ is given by Equation 29.8 as λ = h / p , where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv , where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that λ = h / ( mv ) , or v= 6.63 × 10 –34 J ⋅ s h = = 3.05 × 10 7 m / s –27 –14 mλ 1.67 × 10 kg 1.30 × 10 m c c h h The kinetic energy of the proton is 1 1 KE = 2 mv 2 = 2 (1.67 × 10 –27 kg)(3.05 × 10 7 m / s) 2 = 7.77 × 10 –13 J 26. REASONING According to Equation 27.1, the angle θ that locates the first-order bright fringes (m = 1) is specified by sin θ = λ/d, where λ is the wavelength and d is the separation between the slits. The wavelength of the electron is the de Broglie wavelength, which is given by λ = h/p (Equation 29.8), where h is Planck’s constant and p is the magnitude of the momentum of the electron. 1528 PARTICLES AND WAVES SOLUTION Combining Equations 27.1 and 29.8, we find that the angle locating the firstorder bright fringes is specified by sin θ = λ d = h pd Dividing this result for case A by that for case B, we find sin θ A h / ( pAd ) pB = = sin θ B h / ( pBd ) pA or pB = (1.2 × 10−22 kg ⋅ m/s) sin (1.6 × 10−4 degrees ) = pB = sin ( 4.0 × 10−4 degrees ) pA sin θ A sin θ B 4.8 × 10−23 kg ⋅ m/s 27. REASONING AND SOLUTION The de Broglie wavelength λ is given by Equation 29.8 as λ = h/p, where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv, where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that λ= h mv v= or 6.63 × 10 –34 J ⋅ s h = = 1.41 × 10 3 m / s –27 –9 mλ 1.67 × 10 kg 0.282 × 10 m c c h h 28. REASONING AND SOLUTION We know that λ = h/mv. Solving for the mass yields m= 6.63 × 10 −34 J ⋅ s h = = λv 8.4 × 10 −14 m 1.2 × 10 6 m / s c c h h 6.6 × 10 −27 kg 29. REASONING AND SOLUTION The average kinetic energy of a helium atom is KE = (3/2)kT = (3/2)(1.38 × 10–23 J/K)(293 K) = 6.07 × 10–21 J The speed of the atom corresponding to the average kinetic energy is v= 6 b g= 2 c.07 × 10 J h= 1.35 × 10 2 KE m The de Broglie wavelength is −21 6.65 × 10 −27 kg 3 m/s Chapter 29 Problems λ= 6.63 × 10 −34 J ⋅ s h = = mv 6.65 × 10 −27 kg 1.35 × 10 3 m / s c c h h 1529 7 .38 × 10 −11 m 30. REASONING The de Broglie wavelength λ of a particle is inversely proportional to the h (Equation 29.8), where magnitude p of its momentum, as we see from λ = p h = 6.63×10−34 J·s is Planck’s constant. The electron is moving at a speed v = 0.88c, which is close to the speed of light in a vacuum (c = 3.00×108 m/s). Therefore, we will use mv p= (Equation 28.3) to determine the magnitude of the electron’s relativistic v2 1− 2 c momentum, where m = 9.11×10−31 kg is the electron’s mass. SOLUTION Substituting Equation 28.3 into Equation 9.8, we find that λ= = h h = p mv 2 1− v c2 = h v2 1− 2 mv c 2 6.63 × 10−34 J ⋅ s (9.11×10−31 kg) ( 0.88) (3.00 ×108 m/s)...
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