# I have the question: The speed of a certain proton is 0.45 Mm/s. If the uncertainty in its momentum is to be reduced to 0.0100 percent, what uncertainty in its location must be tolerated?The ans is 0.70 nm . Pls explain the solution

Dear student,

$∆x.∆p\ge \frac{h}{4\mathrm{\pi }}$; where
$∆p=\left(0.01%\right)mv$ ; where m= mass of proton (1.6726 x 10-27Kg) and v = speed of proton (0.45Mm/s), h = planck's constant (6.626 x 10-34 Js)

Regards

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