### Video Transcript

A proton moving through free space
has an uncertainty in its momentum of 4.00 times 10 to the negative 28 kilograms
meters per second. Using the formula Δ𝑥 times Δ𝑝 is
greater than or equal to ℎ over four 𝜋, calculate the minimum possible uncertainty
in the position of the proton. Use a value of 6.63 times 10 to the
negative 34th joule-seconds for the Planck constant. Give your answer to three
significant figures.

In this exercise, we’re told that
we have a proton moving along through free space. This means the proton is moving
with effectively no forces acting on it. Because a proton has mass and
because it’s in motion, it has mass and velocity, and therefore it has momentum. That momentum, though, isn’t known
with complete precision. There’s some uncertainty in it. We’re told a numerical value for
that uncertainty; we can call it Δ𝑝. And using this equation here in our
problem statement, we want to calculate the minimum possible uncertainty in the
position of the proton.

So, here, we know the uncertainty
in the momentum, and we want to calculate the uncertainty in the proton’s
position. We know that Δ𝑥 times Δ𝑝, the
uncertainty in the proton’s position times the uncertainty in its momentum, is
greater than or equal to Planck’s constant divided by four 𝜋. This symbol, greater than or equal
to, means there’s a fundamental limit to how precisely we can know both of these
variables, the position and momentum of the proton, at the same time.

In this exercise, we want to push
this limit as far as we possibly can because we’re calculating the minimum possible
uncertainty in the proton’s position. That minimum occurs when Δ𝑥 times
Δ𝑝 is equal to ℎ over four 𝜋. Remembering that we know Δ𝑝 and we
want to solve for Δ𝑥, we can multiply both sides of our equation by one divided by
Δ𝑝, canceling out Δ𝑝 on the left. And so, now, we have an expression
for the uncertainty in the proton’s position.

To solve for it, all we need to do
is to plug in our given value for ℎ, Planck’s constant, and Δ𝑝, the uncertainty in
the proton’s momentum. With those two values substituted
in, when we calculate this fraction, to three significant figures, we find a result
of 1.32 times 10 to the negative seventh meters. And if we recall the conversion one
nanometer is equal to 10 to the negative ninth meters, then we can write our answer
as 132 nanometers. What we’re saying then is that
there’s some distance in space, we’ve called it Δ𝑥, within which we can say our
proton exists. And the smallest that distance can
be, given our uncertainty in momentum, is 132 nanometers. That’s our minimum position
uncertainty.