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:

$∆x∆p_xâ‰\yen \frac{h}{2}$

Here, $∆x$is the uncertainty in position. $∆p_x$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:

$∆p_x=m∆v_x$

Here, $∆v_x$is the uncertainty in the velocity of the particle.

Rearrange the equation $∆x∆p_xâ‰\yen \frac{h}{2}$for $∆p$.

$∆p_xâ‰\yen \frac{h}{2∆x}$

Substitute $m∆v_x$for $∆p_x$in the above equation and rearrange the equation for $∆v_x$

$$\begin{gathered} m∆v_xâ‰\yen \frac{h}{2∆x} \\ ∆{v_x}_xâ‰\yen \frac{h}{2m∆x} \end{gathered}$$

Substitute $6.626×{10}^{-34}J.s$for $h,50kg$for $m$and $5.0m$for $∆x$in the above equation and calculate the range of velocity.

$$\begin{gathered} ∆v_x=\frac{h}{2m∆x} \\ =\frac{\left(6.626×{10}^{-34}J.s\right)}{2\left(50kg\right)\left(5.0m\right)} \\ =1.3×{10}^{-36}m/s \end{gathered}$$

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