Question Video: Calculating the Minimum Uncertainty of a Proton’s Position Physics • 9th Grade

A proton moving through free space has an uncertainty in its momentum of 4.00 × 10⁻²⁸ kg.m/s. Using the formula Δ𝑥Δ𝑝 ≥ ℎ/4𝜋, calculate the minimum possible uncertainty in the position of the proton. Use a value of 6.63 × 10⁻³⁴ J.s for the Planck constant. Give your answer to 3 significant figures.

02:42

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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.