Chapter 28, Problem 4P

To determine

The accuracy of the electron’s speed.

Answer to Problem 4P

The answer is $\text{5}\text{.8}\text{×}{\text{10}}^{\text{3}}\text{m/s}$

Explanation of Solution

Given info: Mass of electron, $\text{m = 9}\text{.1}1\times {\text{10}}^{-31}\text{kg}$

The uncertainty in the position measurement is $\text{Δx = 2}{\text{.0×10}}^{\text{-8}}\text{m}$

Formula used:

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

Where,

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

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

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

The uncertainty in the momentum is $\Delta \text{p = m }\Delta \text{v}$

Where,

  $v$ is change in velocity.

  $m$ is mass

Calculation:

First calculate the uncertainty in the momentum by using the formula $\Delta \text{x }\ge \frac{h}{\Delta p}$

  $\text{Δp}\ge \frac{\text{h}}{\text{Δx}}\text{=}\frac{\left(\text{1}{\text{.055×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)}{\left(\text{2}{\text{.0×10}}^{\text{-8}}\text{m}\right)}=5.28\times {10}^{-27}\text{kg}\cdot \text{m/s}$

Now, calculate the uncertainty in the velocity by using the formula $\Delta \text{p = m }\Delta v$

  $\text{=}5.28\times {10}^{-27}\text{kg}\cdot \text{m/s}=\left(9.11\times {10}^{-31}\text{kg}\right)\Delta v$

  $\text{Δv}\text{=}\text{5}{\text{.8×10}}^{\text{3}}\text{m/s}$

Conclusion:

The required value is $\text{5}{\text{.8×10}}^{\text{3}}\text{m/s}$ .