Question

If the uncertainty of velocity of a microscopic particle is increase by 25% then calculate the subsequent change in position

Answer 1

Answer:

Explanation:

Use Heisenberg's uncertainty principle,

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

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