If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron



Q.61:- If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/(4π×0.05 nm), is there any problem in defining this value.

Answer:-

Δx = 0.002 nm = 2×10-12 m



Δx × Δp = h/4π
∴ Δp = h/4πΔx = 6.626×10-34 kgm2s-1/(4×3.14×2×10-12 m) = 2.638×10-23 kgms-1
Actual momentum = h/(4π×0.05 nm) = h/(4π×5×10-11 m)
= 6.626×10-34 kgm2s-1/(4×3.14×5×10-11 m) = 1.055×10-24 kgms-1
It cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.