Question

If the position of an electron in a membrane is measured to an accuracy of 1.00 μm, what is the electron’s minimum uncertainty in velocity?

a) 1.05 x 10^3 m/s
b) 2.32 x 10^3 m/s
c) 3.68 x 10^3 m/s
d) 4.94 x 10^3 m/s

Answer 1

According to the uncertainty principle, the minimum uncertainty in velocity of an electron is 5.27 x 10^-25 m/s when the position is measured to an accuracy of 1.00 μm.

Step-by-step explanation:

The uncertainty principle, formulated by Werner Heisenberg, states that it is impossible to simultaneously know the exact position and velocity of a particle. Mathematically, the product of the uncertainties in position (σx) and velocity (σv) is greater than or equal to h/4π, where h is Planck's constant. In this case, the uncertainty in position is given as 1.00 μm. To find the minimum uncertainty in velocity, we can use the equation:

σxσv ≥ h/4π

Substituting the given value of σx and rearranging the equation, we find:

σv ≥ h/4πσx

Plugging in the known values, we get:

σv ≥ (6.63 x 10^-34 J·s)/(4π x (1.00 x 10^-6 m))

σv ≥ 5.27 x 10^-25 m/s