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.
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$\begingroup$ This is my first question . Can some one please answer quickly ? $\endgroup$– TESLAGENCommented Sep 24, 2016 at 16:47
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$\begingroup$ "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? $\endgroup$– SecretCommented Sep 24, 2016 at 16:58
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$\begingroup$ 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. $\endgroup$– TESLAGENCommented Sep 25, 2016 at 9:29
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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.
