Answer:
The minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
Explanation:
Given;
velocity of the proton, v = 2 x 10⁸ m/s
uncertainty in the velocity, = 1 % = 0.01 v = 0.01 x 2 x 10⁸ m/s = 2 x 10⁶ m/s
The momentum of the proton is given by;
ΔP = mv = (1.67 x 10⁻²⁷)(2 x 10⁶ ) = 3.34 x 10⁻²¹ kgm/s
The minimum uncertainty in the position of the proton is given by;
[tex]\delta x = \frac{h}{4\pi \delta P}\\\\ \delta x =\frac{6.626*10^{-34}}{4\pi *3.34*10^{-21}}\\\\ \delta x = 1.578*10^{-14} \ m[/tex]
Therefore, the minimum uncertainty in the position of the proton is 1.578 x 10⁻¹⁴ m
