A proton (rest mass 938.3 MeV/c²)is confined to a small region of space 2.0 nm wide.What is the uncertainty on its speed due to this confinement?

Question

Quizplus · Accepted Answer

The uncertainty in the proton's speed is given by the Heisenberg uncertainty principle:$$\Delta x \Delta p \geq \frac{h}{4\pi}$$where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is the Planck constant.In this case, the uncertainty in position is 2.0 nm, so the uncertainty in momentum is:$$\Delta p \geq \frac{h}{4\pi \Delta x} = \frac{6.626 	imes 10^{-34} 	ext{ J s}}{4\pi 	imes 2.0 	imes 10^{-9} 	ext{ m}} = 2.65 	imes 10^{-24} 	ext{ kg m/s}$$The uncertainty in speed is then:$$\Delta v = \frac{\Delta p}{m} = \frac{2.65 	imes 10^{-24} 	ext{ kg m/s}}{1.673 	imes 10^{-27} 	ext{ kg}} = 1.59 	imes 10^3 	ext{ m/s}$$The closest answer choice is A) 32 m/s, which is incorrect. The correct answer should be 1590 m/s, which is not listed as an answer choice.

A Proton (Rest Mass 938