A proton (mass = 1.67 x 10^-27kg)has a kinetic energy of 1.00 MeV.If its momentum is measured with an uncertainty of 1.00%,what is the minimum uncertainty in its position? (h = 6.63 x 10^-34 J*s and 1 eV = 1.6 x 10^-19 J) A)9.08 x 10^-13 m B)2.28 x 10^-13 m C)9.08 x 10^-14 m D)5.64 x 10^-14 m

Question

Quizplus · Accepted Answer

We can use the Heisenberg Uncertainty Principle: 
Δx * Δp ≥ h/4π 
where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant. 
First, we need to convert the kinetic energy of the proton into momentum: 
K.E. = p²/2m 
1.00 MeV = p²/2(1.67 x 10^-27 kg) 
p ≈ 1.91 x 10^-19 kg m/s 
Now, we can calculate the uncertainty in momentum: 
Δp = 1.00% * p 
Δp ≈ 1.91 x 10^-21 kg m/s 
Finally, we can calculate the minimum uncertainty in position: 
Δx ≥ h/4πΔp 
Δx ≥ (6.63 x 10^-34 J s)/(4π(1.91 x 10^-21 kg m/s)) 
Δx ≥ 2.28 x 10^-13 m 
Therefore, the answer is B, with a minimum uncertainty in position of 2.28 x 10^-13 m.

A Proton (Mass = 1