Calculate the uncertainity in the position `(Delta x)` of an electron if `Delta v` is 0.1%. Take the velocity of electron `=2.2xx10^(6) ms^(-1)` and mass of electron as `9.108xx10^(-31) kg.`

Options

A. `263.47 xx10^(-10) m`
B. `240.46xx10^(10) m`
C. `254.45xx10^(-10) m`
D. `235.33xx10^(-10) m`

Answer

A

Explanation

Given, `Delta v=0.1%` of the velocity of the electron `=(0.1)/(100) xx2.2xx10^(6) =2.2xx10^(3) ms^(-1)`
`Delta x xx m Delta v =(h)/(4 pi)" or "Delta x=(6.63xx10^(-34) Js)/(4xx3.14xx9.108xx10^(-31) kg xx 2.2xx10^(3)" ms"^(-1))=0.02634765 xx 10^(-6) m`
`=263.4765xx10^(-10) m`
Since `Delta x` is much longer than the atomic diameter `(~~ 10^(-10) m)`, the uncertainity principle is applicable in this case.