Chapter 28, Problem 9P

To determine

The uncertainty in position of electron and basketball.

Answer to Problem 9P

The uncertainty in position of electron is $1.4\times {10}^{-3}m.$

The uncertainty in position of basketball is $9.1\times {10}^{-33}m$

The uncertainty for the electron is greater by a factor of $1.5\times {10}^{29}.$

Explanation of Solution

Given info: Velocity is $\text{150}\text{m/s}$

Accuracy is $\text{0}\text{.055\%}$

Mass of ball is $\text{0}\text{.140}\text{kg}$

Mass of electron is $\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}$

Formula used:

It is known that $\text{Δx }\ge \frac{\text{h}}{\text{mΔv}}$

Where,

  $h$ is Planck’s constant, $1.055\times {10}^{-34}J\cdot s$

  $\Delta \text{p}$ is uncertainty in the momentum

  $\Delta v$ is change in velocity.

  $m$ is mass

  $\Delta x$ is the uncertainty in the position measurement

Calculation:

First calculate the uncertainty in the velocity is

  $\text{Δv=}\left(\frac{\text{0}\text{.055}}{\text{100}}\right)\left(\text{150}\text{m}/\text{s}\right)\text{=0}\text{.0825}\text{m}/\text{s}\text{.}$

For the electron, we have

  $\begin{array}{l}\text{Δx}\ge \frac{\text{h}}{\text{m}\text{Δv}}\\ =\frac{\left(1.055\times {10}^{-34}J·s\right)}{\left(9.11\times {10}^{-31}kg\right)\left(0.0825m/s\right)}\\ =1.4\times {10}^{-3}m.\end{array}$

For the baseball, we have

  $\begin{array}{l}\text{Δx}\ge \frac{\text{h}}{\text{m}\text{Δv}}\\ =\frac{\left(1.055\times {10}^{-34}J·s\right)}{\left(0.140kg\right)\left(0.0825m/s\right)}\\ =9.1\times {10}^{-33}m.\end{array}$

After comparing, the uncertainty for the electron is greater by a factor of $1.5\times {10}^{29}.$

Conclusion:

All the required values arecalculated.