If the position o

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