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1. An electron has a speed of500 ms) with uncertainty of 0.02%. What is the uncertainty in locatinglAns. 5-77x104m]its position ? |
Answer» <p>Speed of the electron = v = 500 m/s.uncertainityin speed is 0.02 %∆v = 0.02 % of 500 m/s=0.02100×500= 0.1 m/sAccording to Heisenberguncertainityprinciple,</p><p>∆x.∆p≥h4πwhere∆x = uncertainity in position∆p = uncertainity in momentumh = Planck's constant = 6.63×10^-34kg m2/sto calculate∆p = m∆vm = mass of electron = 9.1×10^-31kg∆v = uncertainity in speed = 0.1 m/s∆p = m∆v =9.1×10^-31kg×0.1 m/s =9.1×10^-32kg m/s∆x.∆p≥h4π∆x≥h4π1∆p∆x≥6.63×10^−34*4×3.1419.1×10^−32= 5.8×10^-4mUncertainity in position =5.8×10^-4m</p> | |