Assume that an electron is moving along an x axis and that you measure its speed to be 2.05xx10^(6)m//s, which can be known with a precision of 0.50%. What is the minimum uncertainty (as allowed by the uncertainty principle in quantum theory) with which you can simultaneously measure the position of the electron along the x - axis ? Calculations : To evaluate the uncertainty Deltap_(x) in the momentum, we must first evaluate the momentum component p_(x). Because the electron's speed v_(x) is much less than the speed of light c, we can evaluate p_(x) with the classical expression for momentum instead of using a relativistic expression. We find p_(x)=mv_(x)=(9.11xx10^(-31)kg)(2.05xx10^(6)m//s) =1.87xx10^(-24)kg.m//s. The uncertainty in the speed is given as 0.50% of the measured speed. Because p_(x) depends directly on speed, the uncertainty Deltap_(x) in the momentum must be 0.50% of the momentum : Deltap_(x)=(0.0050)p_(x) =(0.0050)(1.87xx10^(-24)kg.m//s) =9.35xx10^(-27)kg.m//s Then the uncertainty principle gives us Deltax=(h)/(Deltap_(x))=((6.63xx10^(-34)J.s)//2pi)/(9.35xx10^(-27)kg.m//s) =1.13xx10^(-18)m~~11nm, which is about 100 atomic diameters.