Question

Calculate the uncertainty in the velocity of an electron,if uncertainty in its position is 0.0001m

Answer 1

Use Heisenberg's uncertainty principle,
$\bold{\Delta{x}\times\Delta{P} \geq \frac{h}{4\pi}}$
Here , ∆x is uncertainty in position, ∆P is uncertainty in momentum and hnis planks constant.

Given,
∆x = 0.0001 m
∆P = m∆v , here m is mass of electron
so, m = 9.1 × 10⁻³¹ kg
h = 6.626 × 10⁻³⁴ Js

Now, 0.0001 × 9.1 × 10⁻³¹ ∆v ≥ 6.626 × 10⁻³⁴/4 × 3.14
⇒ 9.1 × 10⁻³⁵ × ∆v ≥ 6.626 × 10⁻³⁴/12.56
⇒∆v ≥ 66.26/(12.56 × 9.1)
⇒ ∆v ≥ 0.58 m

Hence, uncertainty in velocity of electron is 0.58 m