Question

From the relation Δp.Δx ≥ h/4π show that for a particle moving in a circle ΔL.Δθ ≥ h/4π. The quantity ΔL is the uncertainty in the angular momentum and Δθ is the uncertainty in the angle.

Answer 1

"The uncertainty principle delta p delta x greaterthanorequalto h/2pi If a particle is moving with momentum ""p"" on a circle of radius 'r': its angular momentum 'L' is given by L = m v r = (m v) r = pr The Change in L is given as delta L = delta pr implies Therefore delta p delta L/r In moving a distance ""x' on the circle the circle the particle sweeps out an angle ""phi"" in radians Therefore phi = x/r The change in angle is given as delta phi = delta/r implies theta x = delta phi r Substituting the values of delta p & delta x: we get delta p delta x greaterthanorequalto h/4pi (delta L/r) times (delta phi r) greaterthanorequalto h/2pi Cancelling r: we get Delta L delta phi greaterthan h/4pi"