Question

An electron traveling at 4.6×105 m/s has an uncertainty in its velocity of 1.71×105 m/s . What is the uncertainty in its position?

Answer 1

V = 4.6*10^5

uncertainty = 1.71*10^5 m/s

Note that we must use the uncertainty principle of Heisenberg

it involves ONLY uncertainties; that is, if the electron is moving faster or slower, the uncertainty will not depend on that speed

therefore

ΔpΔx≥h4π

dp = mdv

mass of electron = 9.1093*10^-31 kg

then

dp = (9.1093*10^-31)(1.71*10^5) = 1.55769*10^-25

ΔpΔx≥h4π

h = planck constant = 6.626*10 ^-34

ΔpΔx≥h4π

(1.55769*10^-25 )(dx) > (6.626*10 ^-34)(4*3.141592)

solve for dx

dx =  (6.626*10 ^-34)(4*3.141592)/(1.55769*10^-25 ) = 5.3453*10^-8 m