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Suppose that the uncertainty in the momentum of a partícle is equal to this momentum (Ap p). How is the minimum uncertainty in the position of the particle related to its de Broglie wavelength

User Adam Dyga
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1 Answer

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Answer:


\Delta x = (\lambda)/(4\pi)

Step-by-step explanation:

As per de Broglie theory we know that it is given as


P = (h)/(\lambda)

now here we can say that by the principle of uncertainty we have


\Delta x * \Delta P = (h)/(4\pi)

now we can use it to find the uncertainty in position as


\Delta x = (h)/(4\pi \Delta P)

now plug in the value of momentum as per de Broglie theory


\Delta x = (h)/(4\pi ((h)/(\lambda)))


\Delta x = (\lambda)/(4\pi)

So above is the maximum uncertainty in position in terms of de Broglie wavelength

User Steffenhk
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