Answer
The uncertainty in the position of the neutron is Δx≥0.03148nm.
Work Step by Step
Use Heisenberg's uncertainty principle formula:
Δx × Δ(mv)≥h4π
Δx is the uncertainty in position
Δ(mv) is the uncertainty in momentum
h is Planck's constant (h=6.626×10−34J.s)
First Identify the known variables
- Mass of neutron(approximately) : m≈1.675×10^{-27}
- The uncertainty in velocity: Δv=0.01×10^{5}m/s
Because there is only uncertainty in velocity, the uncertainty in momentum is Δ(mv)=mΔv
Next plug in the values into the equation to determine the uncertainty in position of the electron
Δx≥h/4πmΔv=6.626×10^{-34}/4π×(1.675×10^{-27})×(0.01×105)≈3.148×10^{-11}m≈0.03148nm
