$\Delta p \Delta x\geq \frac{h}{4\pi }$

Cecilia Tsai 3C
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

The question in case you were wondering is:
What is the minimum uncertainty in the position of a hydrogen atom in a particle accelerator given that its speed is known to within +/- 5.0 m/s.

Ronald Yang 2F
Posts: 86
Joined: Fri Sep 25, 2015 3:00 am

First, start off with Heisenberg's Indeterminacy Equation:$\Delta p \Delta x\geq \frac{h}{4\pi }$. You can break down Δp into mΔv, since mass is constant, so $m\Delta v \Delta x\geq \frac{h}{4\pi }$. Δx is what you want to find. You can find Δv, which is uncertainty in velocity, by taking the +/- 5 m/s and multiplying it by 2, since the velocity ranges from 5 m/s above and below a certain velocity (not stated). m is what you can calculate with the information "a hydrogen atom." You would do dimensional analysis to yield the mass of that hydrogen atom, so $1 H atom*\frac{1 mol H}{6.02*10^{23} H atoms} * \frac{1.01 g}{1 mol H} * \frac{1 kg}{10^3 g}\approx 1.68 * 10^{-27} kg$. With this mass and Δv, 10 m/s, plug into the equation and solve for Δx to get 3.1*10^-9 m, or 3.1 nm.