Question

You are conducting an experiment in which you can measure the location of individual electron collisions to within 10^-10m. A). What is the theoretical limit to which you can simultaneously measure the momentum of those collisions? Express your answer in kilograms times meter per second to three significant figures.B). What is the uncertainty in the electrons speed? (The electron has a test mass of 9.1 x 10^-31 kg.) Express your answer in meter per second to two significant figures.

Answer 1

Given data:

* The uncertainty in the measure of the location is given as,

$\Delta x=10^{-10}\text{ m}$

Solution:

According to the uncertainty principle, the theoretical limit to measure the momentum is,

$\Delta p\approx\frac{h}{4\pi\Delta x}$

where h is the Planck's constant,

Substituting the known values,

$\begin{gathered} \Delta p\approx\frac{6.626\times10^{-34}}{4\pi\times10^{-10}} \\ \Delta p\approx0.527\times10^{-24}\text{ kg m }s^{-1} \end{gathered}$

Thus, in advanced mathematics, the value of the momentum in kilogram meter per second is,

$\text{0}.527\times10^{-24}\operatorname{kg}ms^{-1}$

(B). The mass of the electron is,

$m=9.1\times10^{-31}\text{ kg}$

Thus, the uncertainty in the electron speed is,

$\Delta v=\frac{\Delta p}{m}$

Substituting the known values,

$\begin{gathered} \Delta v=\frac{0.527\times10^{-24}^{}}{9.1\times10^{-31}} \\ \Delta v=0.058\times10^7\text{ m/s} \\ \Delta v=0.58\times10^6\text{ m/s} \end{gathered}$

Thus, the uncertainty in the electron speed is,

$\Delta v=0.58\times10^6\text{ m/s}$