Question

The velocity of an electron is known to be 1.000 × 105 ms , with an uncertainty of ∆v =

2m 1.00×10 s.

1. (a) What is the minimum uncertainty in the electron’s position, ∆x, in meters?

2. (b) How does this compare to the de Broglie wavelength of the electron?

(c) Your professor (m = 75.0 kg) is pacing in front of the classroom, and you measure their m

velocity to an uncertainty of ∆v = 0.100 s . What is the minimum uncertainty in a measurement of their position?

(d) How does this compare to the height of your professor?

Answer 1

(a).

Heisenberg's Uncertainty principle relates the uncertainty in position () with the uncertainty in momentum () as follows

Furthermore, the uncertainty in momentum can be written as where is the uncertainty in velocity.

Hence,

Given that the uncertainty of electron .

we can calculate the minimum uncertainty in position as follows:

Hence, the minimum uncertainty in position of the electron is about .

(b).

Note that the velocity of the electron is given to be

de-Broglie wavelength of the electron can be calculated as follows:

Hence, the de-Broglie wavelength of the molecule is about .

Hence, the wavelength of the electron is two orders of magnitude smaller than the uncertainty in position. Hence, the position of the electron cannot be measured with precision.

(c).

Given the uncertainty in velocity of the professor = and mass = m = 75.0 kg,

we can calculate the uncertainty in position as follows:

Hence, the minimum uncertainty in measurement of position of the professor is .

(d).

Height of the professor is in the order of 2 m.

Hence, the uncertainty in measurement of the position of the professor is negligible compared to the height and differs by a staggering 36 orders of magnitude.. And hence we can measure the position of the professor with appreciable degree of certainty.