German physicist Werner Heisenberg related the uncertainty of an object’s position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥

Question

German physicist Werner Heisenberg related the uncertainty of an object’s position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck’s constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 1.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?

in progress 0
Bình An 3 years 2021-08-03T10:31:55+00:00 1 Answers 7 views 0

Answers ( )

  1. thongdat2
    0
    2021-08-03T10:33:39+00:00
    Reply

    Answer:

    The uncertainty in the position of the electron is 5.79x10^{-9}m

    Explanation:

    The Heisenberg uncertainty principle is defined as:

    \Lambda p\Lambda x ≥ \frac{h}{4 \pi}  (1)

    Where \Lambda p is the uncertainty in momentum, \Lambda x is the uncertainty in position and h is the Planck’s constant.

    The momentum is defined as:

    p =mv  (2)

    Therefore, equation 2 can be replaced in equation 1

    \Lambda (mv) \Lambda x ≥ \frac{h}{4 \pi}

    Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

    m \Lambda v \Lambda x ≥ \frac{h}{4 \pi} (3)

    Then, \Lambda x can be isolated from equation 3

    \Lambda x ≥ \frac{h}{m \Lambda v 4 \pi}  (4)

    \Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}

    But 1J = Kg.m^{2}/s^{2}

    \Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}

    \Lambda x = 5.79x10^{-9}m

    Hence, the uncertainty in the position of the electron is  5.79x10^{-9}m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )