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## Homework Statement

Show that the smallest possible uncertainty in the position of an electron whose speed is given by [tex]\beta = v/c[/tex] is [tex]\Delta x_{min} = \frac{h}{4 \pi m_0 c}\sqrt{1-\beta^2} [/tex]

## The Attempt at a Solution

Since [tex] \Delta x \Delta p \geq \frac{\hbar}{2} [/tex], we see that [tex] \Delta x_{min} [/tex] occurs when [tex] \Delta p [/tex] has its greatest value.

Relativistically, [tex] \Delta p [/tex] is:

[tex] \Delta p = \Delta ( \frac{m_0 v}{\sqrt{1 - \beta^2}}) = ... = (1-\beta^2)^{-3/2} m_0 \Delta v [/tex]Now the greatest value of [tex] \Delta p [/tex] occurs when [tex] \Delta v [/tex] is c.

Hence

[tex] \Delta x_{min} = \frac{h}{4 \pi m_0 c} (1 - \beta^2)^{3/2} [/tex].

My exponent for gamma is incorrect. Where did I go wrong?

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