A micorscope using suitable photons is employed an electron in an atom within a distance of ` 0.1 A` . What is the uncetrancity involved in the measurment of its velcity ?

Options

A. `5.79xx10^(6) ms^(-1)`
B. `5.79xx10^(7) ms^(-1)`
C. `5.79xx10^(8)ms^(-1)`
D. `5.79xx10^(5)ms^(-1)`

Answer

A

Explanation

(a) By Heisenberg's uncertainty principle
`Deltax xx Delta p_(x) ge (h)/(4pi)`
or ` Delta x xx Delta (mv_(x)) ge (h)/(4pi)`
`Deltax xx Delta_(x) ge (h)/(4 pi m)`
`Deltap` = uncertainty in momentum
`Deltax`=uncertainty in position
`Deltav` = uncertainty in velocity
m= mass of particle
Given that,
`Deltax=0.1Å=0.1xx10^(-10)m`
`m=9.11 xx10^(-31) kg`
h=Planck's constant `=6.626 xx10^(-34) Js`
`pi = 3.4`
Thus,
`Delta v xx 0.1xx10^(-10)=(6.626xx10^(-34))/(4xx3.14xx9.11xx10^(-31))`
`Delta v= (6.626xx10^(-34))/(4xx3.14xx9.11xx10^(-31)xx0.1xx10^(-10))ms^(-1)`
`=5.785xx10^(6) ms^(-1)=5.79xx10^(6) ms^(-1)`