Question

Calculate the minimum uncertainty in velocity of a particle of mass 1.1 x 10 -27 kg if uncertainty in its position is 3 x 10 -10 cm.

Answer 1

Answer:

Uncertainity in velocity = Δv = 1.6 * 10⁻⁴ m/s

Explanation:

$\blacksquare$ We know that, by Heisenbergs Uncertainity Principle

$\implies$ Δx · Δp ≥ $\dfrac{h}{4 \pi}$

$\blacksquare$ We also know that,

$\implies$ Δp = mΔv

$\blacksquare$ Substituting this in the above equation we get,

$\implies$ Δx · mΔv ≥ $\dfrac{h}{4 \pi}$

$\implies$ Δx · Δv ≥ $\dfrac{h}{4 \pi m}$

$\blacksquare$ Given that,

$\implies$ Uncertainity in position = Δx = 3 * 10⁻¹⁰ cm = 3 * 10⁻¹² m

$\implies$ Mass = m = 1.1 * 10⁻²⁷ kg

$\implies$ Planck's constant = h =  J s

$\implies$ Uncertainity in velocity = Δv = ?

$\blacksquare$ Substituting all these values in the equation we get,

$\implies$ 3 * 10⁻¹² * Δv ≥ $\dfrac{h}{4 \pi}$

$\implies$ 3 * 10⁻¹² * Δv ≥  $\dfrac{6.626 * 10^{-34}}{4 * 3.14 * 1.1 * 10^{-27}} = 0.4795 * 10^{27-34}=4.795 * 10^{-8}$

$\implies$ Δv ≥ $\dfrac{4.795 * 10^{-8}}{3 * 10^{-12}}=1.6 * 10^{4} m/s$

Answer 2

Answer:

answer. (by taking approx.)