Question about the Heisenberg Uncertainty Principle

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SUMMARY

The discussion centers on calculating the minimum uncertainty in the speed of a particle using the Heisenberg Uncertainty Principle and de Broglie wavelength. Given an average speed of 4.7 x 105 m/s, the uncertainty in position is set equal to the de Broglie wavelength. By applying the equations Δp * Δy = h/(4π) and λ = h/p, the solution derives that the minimum uncertainty in speed (Δv) is calculated as Δv = v/(4π), leading to a definitive understanding of the relationship between momentum and uncertainty.

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Homework Statement



Suppose the minimum uncertainty in the position of a particle is equal to its de Broglie wavelength. If the particle has an average speed of 4.7 105 m/s, what is the minimum uncertainty in its speed?

Homework Equations



lambda=h/p & (Py)(Y) is greater than or equal to h/4pi

The Attempt at a Solution



I'm not really sure where to start
 
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Let LAMDA = De Broglie Wavelength
DELTAp = uncertainty in linear momentum
DELTAy = uncertainty in position
DELTAv = uncertainty in speed

(Question gives:) DELTAy = LAMDA
Since DELTAp*DELTAy = h/(4pi), then DELTAp*LAMDA = h/(4pi)
Thus, LAMDA = h/(4pi*DELTAp)

Also, LAMDA = h/p

Combine the 2: h/(4pi*DELTAp) = h/p
So: 4pi*DELTAp = p

p = mv, DELTAp = m*DELTAv
4pi*m*DELTAv = mv
4pi*DELTAv = v

We want DELTAv, so:
DELTAv = v/(4pi)

tada.
 
oh man, delta p=m*delta v was the whole key to figuring this out that I NEVER would have thought of. Thank you so much for this detailed solution! You are the master :)
 

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