# 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

$\u2206x.\u2206p\ge \frac{h}{4\mathrm{\pi}}$; where

$\u2206p=(0.01\%)mv$ ; where m= mass of proton (1.6726 x 10

^{-27}Kg) and v = speed of proton (0.45Mm/s), h = planck's constant (6.626 x 10

^{-34 }Js)

$\u2206x=\frac{6.626\times {10}^{-34}}{4\times 3.14\times 0.01\times {10}^{-2}\times 1.6726\times {10}^{-27}\times 0.45\times {10}^{6}}$

$\u2206x=0.701nm$

Regards

**
**