Question

An electron was accelerated through a potential difference of 1.00 ± 0.01 kV. What is the uncertainty of the position of the electron along its line of flight?

Answer 1

If the momentum of a particle is precisely specified, the position is

totally uncertain. This remarkable destruction of one of the fundamentals

                        of classical mechan­ics is a special case of the Heisenberg Uncertainty

                        Principle, It states that if the momentum is known to lie within a range

                        δp, then the position must be uncertain to an extent δz,

                        where

                                                δpδz ≥ (ћ/2)                                                                           (1)

                        Hece ћ = (ћ / 2π) and h = Planck's constant. Note that z is the line of flight.

                        The uncertainty of momentum can be calcu­lated from

                                                δp = m_eδv                                                                              (2)

                        where δv = the uncertainty in velocity. This δv can be calculated from

                        (1/2)m_ev² = eV.

                                    For a potential difference of 0.99 kV,

                                            v = √(2eV / m_e)

                                                = √[{2(1.602 × 10^–19 C) (990V)} / {9.110 × 10^–31 kg}]

                                                = √[{2(1.602 × 10^–19) (990) CV } / {9.110 × 10^–31 kg}]

                                                = √[3.482 × 10¹⁴ (JV^–1V / kg)]

                                                = √[3.482 × 10¹⁴ {(kg m² sec^–2) / kg}]

                                                = √[3.482 × 10¹⁴ m² sec^–2]

                        ∴                  v = 1.866 × 10⁷ m s^–1

                        Similarly, for a potential difference of 1.01 kV,

                                            v = 1.885 × 10⁷ m s^–1

                        Hence      δv = (1.885 – 1.866) × 10⁷ m s^–1

                                                = 0.019 × 10⁷ m s^–1

                                                = 1.9 × 10⁵ m s^–1

                        Solving for δp in equation (2) gives

                                          δp = (9.110 × 10^–31 kg) (1.9 × 10⁵ m s^–1)

                                                = 17.309 × 10^–26 kg m s^–1

                                                = 1.7 × 10^–25 kg m s^–1

                        Rearranging equation (1) to solve for δz gives

                              δz ≥ {(ћ/2) / δp}  = [{1.054 × 10^–34 Js} / {2(1.7 × 10^–25 m s^–1)}]

                                                            = 3.09 × 10^–10 (Js/kg) m s^–1

                                                            = 3.09 × 10^–10 (kg m²/s²) (s/kg) m s^–1

                        or                           δz = 3.09 × 10^–10 m = 309 pm = 3.09  Å.

Hope this helps