Question

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

Answer 1

Given Data :-  L = 10^-10m 

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