s121 - Solutions to Homework#12 Chapter 30 74 Picture the...

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Chapter 30 74. Picture the Problem : When the energy is measured to within a given uncertainty, the time cannot be known to within a greater uncertainty than is stipulated by the Heisenberg uncertainty principle. Strategy: Solve equation 30-20 for the minimum uncertainty in the time. Solution : Calculate the minimum uncertainty in time: ( 29 ( 29 34 13 min 19 max 2 6.63 10 J s 6.6 10 s 2 2 0.0010 eV 1.60 10 J/eV h E t h t E π - - - ∆ ∆ ≥ × × = = = × × Insight: As the energy is measured with greater precision, the time at which the particle had that given energy is known with less certainty. It is impossible to simultaneously measure both energy and time with arbitrary certainty. 78. Picture the Problem : When a proton’s location is known to within an uncertainty of 0.15 nm, it will have a minimum uncertainty in momentum as required by the uncertainty principle. That minimum uncertainty requires the proton to have a nonzero kinetic energy. Strategy: Use equation 30-19 to calculate the minimum uncertainty p in the momentum, and then use p = p to find the kinetic energy of the proton according to the kinetic energy equation 2 ( 2 ). K p m = Solution: 1. (a) Solve equation 30-19 for the minimum uncertainty in momentum: ( 29 34 25 min 9 2 6.63 10 J s 7.0 10 kg m/s 2 2 0.15 10 m h p x h p x - - - ∆ ∆ ≥ × × = = = × × × 2. (b) Calculate the minimum kinetic energy: ( 29 ( 29 ( 29 2 25 2 27 19 7.0 10 kg m/s 0.92 meV 2 2 1.673 10 kg 1.60 10 J/eV p K m - - - × × = = = × × Insight: When we compare the results of this problem with problem 77 (which concerned an electron confined with the same x ), we see that the uncertainty in momentum for the electron and proton are the same. The minimum kinetic energy of the electron, however, is 1840 times greater than the minimum kinetic energy of the proton. Chapter 31 4. Picture the Problem: The image shows an alpha particle approaching a gold nucleus with an initial kinetic energy of 3.0 MeV. The alpha particle comes to rest at a distance d from the gold nucleus when all of the kinetic energy has been converted to electric potential energy. Strategy: equal to the final potential energy: kqQ K d =
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This note was uploaded on 02/10/2012 for the course PHY 102 taught by Professor Korotkova during the Spring '09 term at University of Miami.

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s121 - Solutions to Homework#12 Chapter 30 74 Picture the...

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