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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?

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    $\begingroup$ It's neither; $\Delta$ is the standard deviation. But if you're using the uncertainty principle the way you are, it really doesn't matter since everything is order-of-magnitude anyway. Just choose any reasonable definition you want. $\endgroup$
    – knzhou
    May 14, 2017 at 3:53
  • $\begingroup$ Factors of 2 in HUP: physics.stackexchange.com/q/69604/2451 , physics.stackexchange.com/q/103208/2451 and links therein. $\endgroup$
    – Qmechanic
    May 14, 2017 at 4:31
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    $\begingroup$ @knzhou that should be an answer (although I would then complain that the HUP is a mathematically precise statement about waves and their Fourier transforms, and in that sense is not subject to order-of-magnitude level ambiguity) $\endgroup$
    – David Z
    May 14, 2017 at 6:32

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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.

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