Question

From the Heisenberg uncertainty principle, how would you calculate Δx for an electron with Δv = 0.300 m/s?

Answer 1

`1.95 * 10^(-4)"m"`

Explanation:

I assume that you're familiar with the idea behind the Heisenberg Uncertainty Principle, so I won't go into details about that here.

So, you know that you cannot simultaneously measure with arbitrarily high precision both the position and the momentum of a particle.

More specifically, the uncertainty that you get when measuring its position, `Deltax`, and the ucnertainty that you get when measuring its momentum, `Deltap`, will always satisfy this inequality

`color(blue)(Deltax * Deltap >= h/(4pi))" "`, where

`h` - Planck's constant, equal to `6.626 * 10^(-34)"kg m"^2"s"^(-1)`

Now, a particle's uncertainty in momentum can be written as

`color(blue)(Deltap = m * Deltav)" "`, where

`m` - the mass of the particle.

The mass of the electron is listed as

`m_"electron" = 9.10938356 * 10^(-31)"kg"`

Plug in your values and solve for `Deltax`

`Deltax * m * Deltav = h/(4pi) implies Deltax = h/(4pi) * 1/(m * Deltav)`

`Deltax = (6.626 * 10^(-34)color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)("2"))) color(red)(cancel(color(black)("s"^(-1)))))/(4pi) * 1/(9.01938356 * 10^(-31)color(red)(cancel(color(black)("kg"))) * 0.300color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-1)))))`

`Deltax = color(green)(1.95 * 10^(-4)"m") ->` rounded to three sig figs