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.
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.
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