#### Question

An electron is confined in a box of dimension 1A°. calculate minimum uncertainty in its velocity.

#### Solution

Given Data :- L = 10^{-10}m

Formula :- ∆X_{ma}.∆p_{mi} = ħ

Calculations :- since the electron is probable anywhere within the box of length 10^{-10 }m, the maximum uncertainty in locating it is,

ΔX_{ma }= 10^{-10 }m

ΔX_{ma }.m.ΔV_{mi }= h

`ΔV _(mi) = h/(mΔX_(ma))`

`=(6.63 xx 10^(-34))/(2xx3.14xx9.1xx10^(-31)xx10(-8)`

`= ΔV_(mi) = 1.16xx 10^5 m⁄ sec`

Answer :- minimum uncertainty in velocity = 1.16×10^{5m/s }

Is there an error in this question or solution?

#### APPEARS IN

Solution An electron is confined in a box of dimension 1A°. calculate minimum uncertainty in its velocity . Concept: Applications of Uncertainty Principle.