Question #b47ce

As you know, the Heisenberg Uncertainty Principle states that the position and the momentum of a particle cannot be measured simultaneously with arbitrarily high precision.

In other words, the uncertainty in position, `Deltax`, and the uncertainty in momentum, `Deltap`, must always satisfy the inequality

`color(blue)(ul(color(black)(Deltax * Deltap >= h/(4pi))))`

Here

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

In simple terms, the Heisenberg Uncertainty Principle states that a very precise measurement of a particle's position is accompanied by a very high uncertainty in momentum.

Similarly, a very precise measurement of a particle's momentum is accompanied by a very high uncertainty in position.

Now, the uncertainty in momentum can be calculated by

`color(blue)(ul(color(black)(Deltap = m * Deltav)))`

Here

• `m` is the mass of the proton, listed as `~~ 1.6726 * 10^(-27)"kg"`
• `Deltav` is the uncertainty in velocity

`color(white)(a)`
SIDE NOTE The problem mentions velocity, but that is actually the speed of the proton. I will use speed and velocity interchangeably here, but keep in mind that velocity and speed are not the same thing!
`color(white)(a)`

Now, your proton has a speed of

`v = (1600 +- 55)color(white)(.)"m s"^(-1)`

You can calculate the uncertainty in speed, `Deltav`, by going

`Deltav = v_"max" - v_"min"`

`Deltav = (1600 color(red)(+)55)color(white)(.)"m s"^(-1) - (1600 color(red)(-)55)color(white)(.)"m s"^(-1)`

`Deltav = "110 m s"^(-1)`

The uncertainty in momentum will thus be

`Deltap = 1.6726 * 10^(-27)"kg" * "110 m s"^(-1)`

`Deltap = 1.840 * 10^(-25)"kg m s"^(-1)`

Rearrange the Heisenberg inequality to solve for `Deltax`

`Deltax * Deltap >= h/(4pi) implies Deltax >= 1/(Deltap) * h/(4pi)`

Plug in your values to find

`Deltax >= 1/(1.840 * 10^(-25) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s")))^(-1)) * (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)`

`color(darkgreen)(ul(color(black)(Deltax >= 2.9 * 10^(-10)"m")))`

The answer is rounded to two sig figs.