Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: ( 4π ∆y ) gives ∆p y = h 6.63 × 10−34 J ⋅ s = = 1.8 × 10−20 kg ⋅ m/s ( 3.0 × 10−15 m ) 4π ∆y 4π 40. REASONING Suppose the object is moving along the +y axis. The uncertainty in the object’s position is ∆y = 2.5 m. The minimum uncertainty ∆py in the object’s momentum is specified by the Heisenberg uncertainty principle (Equation 29.10) in the form (∆py)(∆y) = h/(4π). Since momentum is mass m times velocity v, the uncertainty in the velocity ∆v is related to the uncertainty in the momentum by ∆v = (∆py)/m. SOLUTION a. Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows: h ∆ p y ( ∆y ) = 4π ( ) ∆ py = h 6.63 ×10−34 J ⋅ s = = 2.1×10−35 kg ⋅ m/s 4π ∆y 4π ( 2.5 m ) b. For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by ∆ p y 2.1×10−35 kg ⋅ m/s ∆v y = = = 4.7 ×10−34 m/s m 0.045 kg 1536 PARTICLES AND WAVES c. For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given by ∆ p y 2.1× 10−35 kg ⋅ m/s ∆v y = = = 2.3 × 10−5 m/s −31 m 9.11× 10 kg 41. SOLUTION The minimum uncertainty ∆y in the position of the particle is related to the minimum uncertainty ∆ p y in the momentum via the Heisenberg uncertainty principle. To cast this relationship into a form that gives us the desired percentage for the minimum uncertainty in the speed, we note that the minimum uncertainty in the position is specified as the de Broglie wavelength λ. We can then express the de Broglie wavelength in terms of Planck’s constant h and the magnitude py of the particle’s momentum. The magnitude of the momentum is related to the mass m and the speed vy of the particle. SOLUTION The percentage minimum uncertainty in the speed is Percentage = ∆v y vy ×100% (1) According to the Heisenberg uncertainty principle, the minimum uncertainty ∆ p y in the momentum and the minimum uncertainty ∆y in the position of the particle are related according to h (29.10) ∆ p y ( ∆y ) = 4π ( ) We know that ∆y is equal to the de Broglie wavelength λ = h / p y (Equation 29.8), where the magnitude of the momentum is p y = mv y (Equation 7.2). Thus, we have ∆y = λ = h h = p y mv y Substituting this result for ∆y into Equation 29.10, we obtain ( ∆ py h mv y ) ( ∆y ) = ( ∆ p y ) h = 4π (2) The last step in our transformation of the uncertainty principle is to realize that () ∆ p y = ∆ mv y = m∆v y , since the mass is constant. Substituting this expression for ∆ p y into Equation (2) shows that Chapter 29 Problems h ( ∆ p y ) mv y = m∆v y ( h ) mv y h = 4π ∆v y or vy = 1537 1 4π Using this result in Equation (1), we find that Percentage = ∆v y vy ×100% = 1 ×100% = 8.0% 4π 42. REASONING The mass m of the particle is related to its rest energy E0 by E0 = mc2 (Equation 28.5). Therefore, if there is a minimum uncertainty ∆E0 in measuring the rest energy of the particle, there will be a corresponding uncertainty ∆m in measuring its mass: ∆E0 = ( ∆m ) c 2 ∆m = or ∆E0 (1) c2 The minimum uncertainty ∆E0 in the particle’s rest energy is related to the length of time ∆t h the particle exists in a state by the Heisenberg uncertainty principle: ( ∆E0 ) ( ∆t ) = 4π (Equation 29.11), where h = 6.63×10−34 J·s is Planck’s constant. SOLUTION Solving Equation 29.11 for the minimum uncertainty ∆E0, we obtain ∆E0 = h 4π ( ∆t ) (2) Substituting Equation (2) into Equation (1), we find that ∆m = ∆E0 c 2 = h 6.63 ×10−34 J ⋅ s = 4π ( ∆t ) c 2 4π 7.4 ×10−20 s 3.00 ×108 m/s ( )( ) 2 = 7.9 ×10−33 kg 1538 PARTICLES AND WAVES 43. SSM REASONING In order for the person to diffract to the same extent as the sound wave, the de Broglie wavelength of the person must be equal to the wavelength of the sound wave. SOLUTION a. Since the wavelengths are equal, we have that λ sound = λ person λ sound = h mperson v person Solving for v person , and using the relation λ sound = v sound / f sound (Equation 16.1), we have v person = = h mperson ( v sound / f sound ) = h f sound m person v sound ( 6.63 × 10 –34 J ⋅ s)(128 Hz ) = ( 55.0 kg)(343 m / s) 4 .50 × 10 –36 m / s b. At the speed calculated in part (a), the time required for the person to move a distance of one meter is t= x 1.0 m = v 4.50 × 10 –36 m / s 1 F1.0 h I F day I F 1 year I = G s JG h JG days J H KH 2KH 3600 24.0 365.25 K 1444444 4444443 7.05 × 10 27 years Factors to convert seconds to years 44. REASONING The energy of a photon is related to its frequency and Planck’s constant. The frequency, in turn, is related to the speed and wavelength of the light. Thus, we can relate the energy to the wavelength. The given relationship between the wavelengths will then allow us to determine the unknown energy. SOLUTION The energy E of a photon with frequency f is E = hf (29.2) where h is Planck’s constant. The frequency is related to the speed c and wavelength λ of the light according to c (16.1) f= λ Chapter 29 Problems 1539 Substituting this ex...
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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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