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  • 3 years ago

Why is \(\Delta q = \lambda\)? Let's say I'm measuring an electron's position with light. I agree that \(\Delta p = h/\lambda\), where \(\lambda\) is the wavelength of light that I use. But how does that lead to the HUP?

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  1. vf321
    • 3 years ago
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    experimentX's link did the job.

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