An electron is fired at a speed of 100 kms-1, from a distance of 1 m, towards the center of a...

Question:

An electron is fired at a speed of 100 km{eq}\cdot{/eq}s{eq}^{-1}{/eq}, from a distance of 1 m, towards the center of a target. Estimate the maximum possible accuracy that can be achieved.

Velocity:

The term 'velocity' can be defined as the change in the position of the object or the particle with respect to the time. Velocity can be calculated in meters per second.

Answer and Explanation:


We are given the following data:

  • The speed of the electron is {eq}v = 100\,{\rm{km}} \cdot {{\rm{s}}^{ - 1}} = 100000\,{\rm{m/s}} {/eq}
  • The distance is {eq}d = 1\,{\rm{m}} {/eq}


The expression for the momentum is given by:

{eq}\Delta p = m \times v {/eq}, where:

  • {eq}m = 9.1 \times {10^{ - 31}}\,{\rm{kg}}{/eq} is the mass of the electron.


Substituting the values into the above equation, we have:

{eq}\begin{align*} \Delta p &= m \times v\\ \Delta p &= \left( {9.1 \times {{10}^{ - 31}}} \right)\left( {100000} \right)\\ \Delta p &= 9.1 \times {10^{ - 26}}\,{\rm{kg - m/s}} \end{align*}{/eq}


The expression for the possible accuracy from the Heisenberg uncertainty principle is given by:

{eq}\Delta x \cdot \Delta p = \dfrac{h}{{2\pi }} {/eq}, where:

  • {eq}h = 6.626 \times {10^{ - 34}} {/eq} is Planck's constant


Substituting the values in the above equation gives us:

{eq}\begin{align*} \Delta x \cdot \Delta p &= \dfrac{h}{{2\pi }}\\ \Delta x \cdot \left( {9.1 \times {{10}^{ - 26}}} \right) &= \dfrac{{6.626 \times {{10}^{ - 34}}}}{{2\pi }}\\ \Delta x &= 1.148\, \times {10^{ - 9}}\,{\rm{m}}\\ \Delta x &= \boxed{1.148\,{\rm{nm}}} \end{align*}{/eq}


Thus, the maximum possible accuracy achieved is Δx = 1.148 nm.


Learn more about this topic:

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Linear Velocity: Definition & Formula

from MCAT Prep: Help and Review

Chapter 14 / Lesson 12
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