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 , is there any problem in defining this value.

By referring to Heisenberg’s uncertainty principle, we know

∆x × ∆p = h/4∏


Where,


∆x = uncertainty in position of the electron


∆p = uncertainty in momentum of the electron


∆x = 0.002nm = 2× 10-12m [given]


Finding the value of ∆p:


∆p = [h/4π] × [1 / ∆x]


∆p = [1/0.002] × {6.626× 10-34/4× [3.14]}


= [1 / 2× 10-12] × {6.626×10-34/4× 3.14}


= [5.2728×10-35] × [5×1011]


= 2.6364 × 10-23 Jsm-1


∆p = 2.637 × 10-23 kgms-1 {1 J – 1 kgms2s-1}


Actual momentum = h / [4n× 5× 10-11]


= [6.626× 10-34] / [4× 3.14× 5× 10-11]


= 1.055 × 10-24 kg m/sec


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