Question

Quantum Mechanics

The largest known atom, francium, has an atomic diameter of 540 pm. What is the uncertainty in momentum, ∆p, and the resulting uncertainty in velocity, ∆v, if ∆x = 540 pm for an electron in Fr?
Answer:
∆p ≥ 9.77 × 10⁻²⁶ kg · m/s; ∆v ≥ 1.07 × 10⁵ m/s

Answer 1

As per Heisenberg uncertainity principle

The product of uncertainities of momentum and position is greater or equal to h /4π

∆p X ∆x ≥ h /4π

∆p ≥ h /4π∆x

h = 6.62 X 10^-34 J s

π = 3.14

∆x = 540 pm = 540 X 10^-12 m

∆p ≥ 6.62 X 10^-34 / 4 X 3.14 X 540 X 10^-12

∆p ≥ 9.77 X 10^-26 Kg .m/s

b) now

∆p ≥ 9.77 X 10^-26 Kg .m/s

Momentum = Mass X velocity

uncertainity in momentum = Mass of electron X uncertainity in velocity

m∆v ≥ 9.77 X 10^-26 Kg .m/s

Mass of electron = 9.1 X10^-31 Kg

∆v ≥ 9.77 X 10^-26 / mass

∆v ≥ 9.77 X 10^-26 / 9.1 X10^-31 = 1.07 X 10^5 m /s (uncertainity in velocity)