91Ó°ÊÓ

The given data can be listed below as,

• The value of the Plank’s constant is,$h=0.60\text{J}·\text{s}$ .
• The mass of the baseball is,$m=0.50\text{kg}$ .
• The speed of the baseball is, $v=20\text{m/s}$.
• The uncertainty in the speed is, $\mathrm{Δ}v=1.0\text{m/s}$.

In this question, the value of the uncertainty in the baseball position can be determined with the help of the Heisenberg uncertainty principle. The relation between the uncertainty in position and speed is an inverse one.

The expression to calculate the uncertainty in the position of the baseball is expressed as,

$\begin{array}{rcl}\left(\mathrm{Δ}x\right)\left(\mathrm{Δ}p\right)& =& \frac{h}{2\mathrm{π}}\\ \mathrm{Δ}x& =& \frac{h}{2\mathrm{π}\mathrm{Δ}p}\\ & =& \frac{h}{2\mathrm{π}m\mathrm{Δ}v}\end{array}$

Here, $\mathrm{Δ}x$i s the uncertainty in the position of the baseball, $\mathrm{Δ}p$is the uncertainty in the momentum of the baseball.

Substitute all the known values in the above equation.

$\begin{array}{rcl}\mathrm{Δ}x& =& \left[\frac{0.60\text{J}·\text{s}}{2\mathrm{π}\left(0.50\text{kg}\right)\left(1.0\text{m/s}\right)}\right]\\ & ≈& 0.191\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}\\ & ≈& \left(0.191\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}\right)\left(\frac{1\text{m}}{1\text{J}·{\text{s}}^{\text{2}}\text{/kg}·\text{m}}\right)\\ & ≈& 0.191\text{m}\end{array}$

Thus, the uncertainty in the position of the baseball is 0.191 m.