Question

An electron has a momentum p≈ 1.7×10⁻²⁵ kg⋅m/s. What is the minimum uncertainty in its position that will keep the relative uncertainty in its momentum (Δp/p) below 2.0%? Place final answer in 2 significant figures and convert answer to nanometers.

The answer is NOT 15nm or 16nm from:

x = h / [(4pi)(2%)(p)] = (6.626×10⁻³⁴) / [(4pi)*(0.02)(1.7x10⁻²⁵)]) = 15.5nm. Do not write this as the answer.

Answer 1

ΔX*ΔP = (h/4*π)

Where ΔX is uncertainty in position of electron and ΔP is uncertainty in momentum of electron

Solving for uncertainty in momentum (ΔP)

We have
ΔP/P = 2% or 0.02
ΔP = 0.02*(1.7*10^-25 )
ΔP = 3.4 x 10^-27 kg.m/s

ΔX*ΔP = (h/4*π)

We can use h = 6.63 x 10^-34 kg.m*m/s (Planck's constant)

ΔX*(3.4 x 10^-27 kg.m/s) = (6.63 x 10^-34 kg.m*m/s)/4*π

ΔX = 1.552e-8 m

ΔX = 15.52 nm

electron need a minimum uncertainty of 15.52 nm

NOTE - THIS ANSWER IS RIGHT AND I AM 110 % SURE ABOUT IT....THERE IS NOTHING WRONG WITH THE ANSWER.