Question

The uncertainty in the momentum of electron is 1 x 10⁻⁵ kg m/s.The minimum uncertainty in its position will be

1. 1.05 x 10⁻²⁸ m
2. 1.05 x 10⁻²⁶ m
3. 5.27 x 10⁻³⁰ m
4. 5.25 x 10⁻²⁸ m

Show the calculations.Lazy answers will be deleted

Answer 1

heya!!!

we know that ∆x.∆p≥h/4π

→1×10^-5×∆x≥6.6×10^-34/4×3.14

→∆x≥6.6×10^-34/(4×3.14)(10^-5)

→∆x≥0.525×10^-29

→∆x≥5.25×10^-30m

the minimum value is 5.25×10^-30m

→the correct option →3

I hope this will help u ;)

Answer 2

Correct option - C) 5.27×10^-30 m