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The uncertainty principle states that $${\Delta}x{\Delta}p{\geq}\frac{h}{4\pi}.$$

In experimental uncertainty, ${\Delta}x$ is actually half the range (since the value implies that the actual value might be ${\Delta}x$ greater or ${\Delta}x$ less).

In this equation, is ${\Delta}x$ (or ${\Delta}p$) given by the range, or half the range just like experimental uncertainties? Specifically, in a hydrogen atom, is ${\Delta}x$ of the orbiting electron given by the diameter of the orbit or the radius of the orbit, and is ${\Delta}p$ given by twice its magnitude of momentum?

Qmechanic
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1 Answers1

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$\Delta x$ is neither the radius nor the diameter of your orbit. It is the standard deviation representing the region in which the particle could be.