Calculate the uncertainty in the position `(Delta x)` of an electron if `Delta v is 0.1 %` .Take the velocity of electron `= 2.2 xx10^(6) ms^(-1)` and mass of electron as `9.108 xx 10^(-31)kg`

Explanation

Gives `Delta v = 0.1 % `of the velocity of the electron
`= (0.1)/(100) xx 2.2 xx 10^(6) m s^(-1)`
`Delta x xx m Delta v = (h)/(4pi)`
or `Delta x = (6.63 xx 10^(-34) Js)/( 4 xx 3.14 xx 9.108 xx 10^(-31) kg xx 2.2 xx 10^(2) ms^(-1))`
`= 0.02624765 xx 10^%(-6) m`
`= 262.4765 xx 10^(-10) m`
Since `Delta x ` is much longer than the atomic diameter `(+10^(-10)m`1 the uncertainty principle is applicable in this case.