8 27.1 The Heisenberg uncertainty principles Example 4: Minimum energy of a particle in a box -zero point energy Solution: For a particle in a box of length L (4p)2=(p-p) (p2-2p-p2) P-p=p h (4p)2=p2≥( The microscopic In particle cannot E ≠0 be rest! 2m 2mL s 27.1 The Heisenberg uncertainty principles Example 5: a proton is known to be in the nucleus of an atom. The size of the nucleus is about 1.0x10- 4m. Find a Pmin=?; b vmin=? For nonrelativistic situation c is it reasonable to express its momentum in classical expression? Solution: a. According to the uncertainty principle h6.626×10 4p 10×104=6.6×10-20kgm/s b The minimum speed of the proton 6.6×10 40×107m/s 1.67×10 66 Example 4: Minimum energy of a particle in a box —zero point energy 2 2 2 2 2 2 2 ( 2 ) ( ) ( ) p p p p pp p p p p av av = − = = − − ∆ = − 0 2 2 ( ) ( ) 2 2 2 2 2 2 = ≥ ≠ = ≥ mL h m p E x h p p ∆ ∆ For a particle in a box of length L §27.1 The Heisenberg uncertainty principles Solution: The microscopic particle cannot be rest! Example 5: A proton is known to be in the nucleus of an atom. The size of the nucleus is about 1.0×10-14 m. Find a. pmin=?; b. vmin=? For nonrelativeistic situation. c. is it reasonable to express its momentum in classical expression? Solution: a. According to the uncertainty principle 6.6 10 kg m/s 1.0 10 6.626 10 20 14 34 = × ⋅ × × = = − − − x h p ∆ ∆ b. The minimum speed of the proton 4.0 10 m/s 1.67 10 6.6 10 7 27 20 min = × × × = = − − m p v §27.1 The Heisenberg uncertainty principles