In an atom, an electron is moving with a speed of ` 600 m//s` with an accuracy of ` 0.005%` . Certainty with which the position of the electron can be localized is :
` ( h = 6.6 xx 10^(-34) kg m^2 s^(-1)` ,
mass of electron ` (e_m) = 9. 1 xx10^(-31) kg)`.

Options

A. `1.92 xx 10^(-3)m`
B. `3.84 xx 10^(-3)m`
C. `5.10xx10^(-3)m`
D. `1.52xx10^(-4)m`

Answer

A

Explanation

Uncertainty in velocity `(DeltaV) = (0.005)/(100)(600ms^(-1))`
`= 0.030ms^(-1)`
To calculate the incertainty in position `(Deltax)`, we apply Heisenberg's uncertainty principle:
`(Deltax)(mDeltav_(x)) ge (h)/(4pi)`
or `Deltaxge (h)/(4pimDeltav_(x))`
`ge((6.6 xx 10^(-34)Js))/(4(3.14)(9.1xx10^(-31)kg)(0.030ms^(-1)))`
`ge 1.9 xx 10^(-3) m`