91Ó°ÊÓ

Heisenberg uncertainty principle states that it is not possible to make a simultaneous determination of the position and the momentum of a particle with unlimited precision. The mathematical representation of Heisenberg uncertainty principle is as follows:

$\mathrm{Δ}x\mathrm{Δ}{p}_xâ‰\yen \frac{h}{2}$

Here, $\mathrm{Δ}x$is the uncertainty in position, $\mathrm{Δ}{p}_x$is the uncertainty in momentum, and$h$ is the Planck's constant.

The uncertainty in momentum of a particle is also given by the following relation:

$\mathrm{Δ}{p}_x=m\mathrm{Δ}{v}_x$

Here,$\mathrm{Δ}{v}_x$ is the uncertainty in the velocity of the particle.

Rearrange the equation $\mathrm{Δ}x\mathrm{Δ}{p}_xâ‰\yen \frac{h}{2}$for $\mathrm{Δ}{p}_x$.

$\mathrm{Δ}{p}_xâ‰\yen \frac{h}{2\mathrm{Δ}x}$

Substitute $m\mathrm{Δ}{v}_x$for $\mathrm{Δ}{p}_x$in the above equation and rearrange the equation for $\mathrm{Δ}{v}_x$.

$\begin{array}{l}m\mathrm{Δ}{v}_xâ‰\yen \frac{h}{2\mathrm{Δ}x}\\ \mathrm{Δ}{v}_xâ‰\yen \frac{h}{2m\mathrm{Δ}x}\end{array}$

Substitute $6.626×{10}^{−34}\text{J}â‹\dots \text{s}$for $h,50\text{kg}$for $m$and $5.0\text{m}$for $\mathrm{Δ}x$in the above equation and calculate the range of velocity.

$\begin{array}{l}\mathrm{Δ}{v}_x=\frac{h}{2m\mathrm{Δ}x}\\ =\frac{\left(6.626×{10}^{−34}\text{J}â‹\dots \text{s}\right)}{2(50\text{kg})(5.0\text{m})}\\ =1.3×{10}^{−36}\text{m}/\text{s}\end{array}$

Hence, she cannot be sure that she is at rest and her velocity is likely to be within the range of $1.3×{10}^{−36}\text{m}/\text{s}$.