s 27. 1 The Heisenberg uncertainty principles Example 2: You measure the diameter of a shotgun pellet of mass 1.0 g to be 1.00+0.01 mm with a micrometer. What is the minimum uncertainty of its momentum if the magnitude of its momentum is measured simultaneously? Solution: h6.626×10 4 =7×10-29kgm/s 1.0×10 pp=m,v=10×10-×100=0, 1kg msY It is exceedingly small. As a result, for such a macroscopic particle, the uncertainty principle has no effect 8 27.1 The Heisenberg uncertainty principles Example 3: The accelerating potential difference of TV kinescope is 9kv, the diameter of exit of electron gun is 0. 1 mm, is it reasonable to describe the electrons as classical particles? Solution: The uncertainty of the speed A4=4x·m4v≥h h 6.626×10-34 ν =727m/s mx9.1×10-×0.1×10 The speed of the electrons =,2AE=2e4=2×1.6×10°×103 ≈56×107m/s 9.11×105 Example 2: You measure the diameter of a shotgun pellet of mass 1.0 g to be 1.00±0.01 mm with a micrometer. What is the minimum uncertainty of its momentum if the magnitude of its momentum is measured simultaneously? Solution: 7 10 kg m/s 1.0 10 6.626 10 29 5 34 = × ⋅ × × = = − − − x h px ∆ ∆ It is exceedingly small. As a result, for such a macroscopic particle, the uncertainty principle has no effect. §27.1 The Heisenberg uncertainty principles 3 1 p p 1.0 10 100 0.1 kg m s − − p = m v = × × = ⋅ ⋅ Example 3: The accelerating potential difference of TV kinescope is 9kV, the diameter of exit of electron gun is 0.1 mm, is it reasonable to describe the electrons as classical particles? Solution: 7.27m/s 9.11 10 0.1 10 6.626 10 31 3 34 = × × × × = = = ⋅ ≥ − − − m x h v x p x m v h x x x ∆ ∆ ∆ ∆ ∆ ∆ The uncertainty of the speed The speed of the electrons 5.6 10 m/s 9.11 10 2 2 2 1.6 10 10 7 31 19 3 ≈ × × × × × = = = − − m e V m KE v ∆ §27.1 The Heisenberg uncertainty principles