1140_Physics ProblemsTechnical Physics

1140_Physics ProblemsTechnical Physics - Chapter 40 Section...

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Chapter 40 481 Section 40.8 The Uncertainty Principle P40.48 (a) ∆∆ px mvx =≥ = 2 so v h mx ≥= = 4 2 42 0 0 1 0 0 0250 π Js kg m ms .. . bg a f . (b) The duck might move by 025 5 125 ms s m af = . With original position uncertainty of 1.00 m, we can think of x growing to 100 225 . m m m += . P40.49 For the electron, pmv e == × ×= × −− 9 11 10 500 1 00 10 4 56 10 31 4 32 . kg m s kg m s ej x h p ×⋅ = 4 6626 10 116 34 32 . . 44 . 5 61 0 k gm s mm . For the bullet, × = × 0 020 0 500 1 00 10 1 00 10 43 . kg m s kg m s b g x h p × 4 528 10 32 . m . P40.50 y x p p y x = and dp h y ∆≥ 4 . Eliminate p y and solve for x . xp y d h x = 4 : x × × 4 1 00 10 100 1 00 10 200 10 32 3 34 . . kg m s m m The answer, x 379 10 28 m , is 190 times greater than the diameter of the Universe! P40.51 With x 21 0 15 m m, the uncertainty principle requires p x x × = 2 26 10 20 k g m s . The average momentum of the particle bound in a stationary nucleus is zero. The uncertainty in momentum measures the root-mean-square momentum, so we take p rms ≈× 31 0 20 kg m s . For an electron, the non-relativistic approximation pm v e = would predict v 0 10 ms, while v cannot be greater than c . Thus, a better solution would be Em c p c m c ee =+ L N M O Q P ≈= 2 2 2 12 2 56 MeV γ
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