Physics Solution Manual for 1100 and 2101

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Unformatted text preview: mA + mB)vf, where vf is the common speed of the two particles. We can evaluate this momentum by using the law of conservation of momentum, which indicates that the total momentum after the collision is the same as it is before the collision. Before the collision only particle B is moving, so that the magnitude of the total momentum at that time has a value of mBv0B, where v0B is the initial speed of particle B. Assuming the particles travel along the +x axis, we write the conservation of linear momentum as follows: + ( mA + mB ) vf = +mBv0B 14243 14 244 4 3 Total momentum after collision Total momentum before collision Using this result, we find that the desired de Broglie wavelength is λf = h h h = = pf ( mA + mB ) vf mBv0B Chapter 29 Problems 1533 But the term on the far right is just the given de Broglie wavelength of the incident particle B. Therefore, we conclude that λf = 2.0 × 10−34 m . 35. SSM REASONING The de Broglie wavelength λ of the electron is related to the magnitude p of its momentum by λ = h/p (Equation 29.8), where h is Planck’s constant. If the speed of the electron is much less than the speed of light, the magnitude of the electron’s momentum is given by p = mv (Equation 7.2). Thus, the de Broglie wavelength can be written as λ = h/(mv). When the electron is at rest, it has electric potential energy, but no kinetic energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is 1 mv 2 . The 2 conservation of energy states that the final total energy of the electron equals the initial total energy: 1 mv 2 = eV 2 2 1 24 Initial 3 4 3 1 total Final total energy energy Solving this equation for the final speed gives v = 2eV / m . Substituting this expression for v into λ = h/(mv) gives λ = h / 2meV . SOLUTION After accelerating through the potential difference, the electron has a de Broglie wavelength of λ= h 2meV = 2 ( 9.11 × 10 6.63 × 10−34 J ⋅ s −31 kg ) (1.60 × 10 −19 C ) ( 418 V ) = 6.01 × 10−11 m 36. REASONING AND SOLUTION The energy of the photon is E = hf = hc/λphoton , while the kinetic energy of the particle is KE = (1/2)mv2 = h2/(2mλ 2). Equating the two energies and rearranging the result gives λphoton /λ = (2mc/h)λ. Now the speed of the particle is v = 0.050c, so λ = h/(0.050 mc), and λphoton /λ = 2/0.050 = 4.0 × 10 1 37. SSM WWW REASONING AND SOLUTION According to the uncertainty principle, the minimum uncertainty in the momentum can be determined from ∆p y ∆y = h / ( 4π ) . 1534 PARTICLES AND WAVES Since p y = mv y , it follows that ∆p y = m∆v y . Thus, the minimum uncertainty in the velocity of the oxygen molecule is given by ∆v y = h 6.63 × 10 –34 J ⋅ s = = 8.3 × 10 –6 m/s 4π m ∆y 4π 5.3 × 10 –26 kg 0.12 × 10 –3 m ( )( ) 38. REASONING When particles pass through the slit the great majority fall on the screen between the first dark fringes on either side of the central bright fringe. The first dark fringes are located by the angles −θ (below the midpoint) and +θ (above the midpoint), so this is the minimum range of angles over which the particles spread out. As Figure 29.15 ∆p y shows, the angle θ is found from θ = tan −1 (Equation 1.4), where ∆py is the p x uncertainty in the y component py of a particle’s momentum, and px is the x component of a particle’s momentum. The given de Broglie wavelength λ = 0.200 mm of the particles is the wavelength they possess before passing through the slit, when their momentum has only the x component px. Therefore, the x component of each particle’s momentum is given by h px = (Equation 29.8), where h = 6.63×10−34 J·s is Planck’s constant. λ We are given that the uncertainty in the position of each particle along the y direction is equal to one-half the width of the slit: ∆y = 1 W . The minimum uncertainty in the y 2 component ∆py of a particle’s momentum, then, is found from the Heisenberg uncertainty principle: h ∆p y ( ∆y ) = (29.10) 4π () SOLUTION Solving Equation 29.10 for ∆py and substituting ∆y = 1 W , we obtain 2 ∆p y = h h h = = 4π ( ∆y ) 4π 1 W 2π W 2 Substituting Equation (1) and px = we find that ( ) ∆p y (Equation 29.8) into θ = tan −1 p λ x h (1) (Equation 1.6), Chapter 29 Problems h ∆p y 2π W θ = tan −1 = tan −1 p h x λ = tan −1 λ 2π W 1535 633 × 10 −9 m = tan −1 −3 2π ( 0.200 × 10 m ) = 0.0289o Therefore, the particles spread out over the range −0.0289° to +0.0289° . 39. SSM WWW REASONING The uncertainty in the electron’s position is ∆y = 3.0 × 10−15 m. The minimum uncertainty ∆py in the y component of the electron’s momentum is given by the Heisenberg uncertainty principle as ∆p y = h / ( 4π ∆y ) (Equation 29.10). SOLUTION Setting ∆y = 3.0 × 10−15 m in the relation ∆p y = h /...
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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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