# How do you solve for the minimum kinetic energy given that you have an area of movement of the particle?

How do you solve for the minimum kinetic energy given that you have an area of movement of the particle? It seems kind of random to me. I thought of using the momentum given that you have the position uncertainty and then calculating the lowest possible momentum from the deviation by subtracting it from the expected momentum . And then, we square and divide by twice the mass of the particle.

• This is my first question . Can some one please answer quickly ? Commented Sep 24, 2016 at 16:47
• "using the momentum given that you have the position uncertainty" Are you talking about expressing $\Delta p$ in terms of $\Delta x$ using the uncertainty principle? What exactly is the area of movement, is it $(\Delta x)^2$. Perhaps the question can be made clearer with some equations to elaborate? Commented Sep 24, 2016 at 16:58
• Yes precisely, I'm talking about calculating the uncertainty in p given ∆x , and the area of movement I'm concerned with is the ∆x , due to the fact that the particle can move through a given area only. Commented Sep 25, 2016 at 9:29

First, consider the uncertainty principle:$$\Delta p \Delta x \ge \frac{\hbar}{2}$$ If you plug in the uncertainty in position $\Delta x$ you should be able to get the uncertainty in momentum $\Delta p$: $$\Delta p \ge \frac{\hbar}{2* \Delta x}$$ This says that the momentum is at minimum $\frac{\hbar}{2*\Delta x}$, but the inequality says that the uncertainty in momentum is much higher.