Chapter 28.2, Problem 20SSC

To determine

To Explain:A tremendous amount of energy is required when an electron cloud is reduced to almost the size of the nucleus, using Heisenberg uncertainty principle.

Explanation of Solution

Introduction:

As per Heisenberg’s uncertainty principle, the position and velocity of an entity cannot be determined exactly at the same time.

Electron cloud is a region around nucleus where electrons are found. If the electron cloud is reduced to a size of nucleus, the uncertainty of finding the electron reduces which increases the uncertainty in its velocity i.e. momentum. This means, a lot of energy would be needed to reduce the size of the electron cloud to the size of the nucleus.

Size of the nucleus is $\Delta x={10}^{-14}\text{m}$ .

As per Heisenberg’s uncertainty principle, $\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$

Here, h is the Planck’s constant.

  $\begin{array}{l}\Delta p\ge \frac{h}{4\pi \left(\Delta x\right)}\\ \Delta p\ge \frac{6.63\times {10}^{-34}\text{J}\cdot \text{s}}{4\pi \left({10}^{-14}\text{m}\right)}\\ \Delta p\ge 5.28\times {10}^{-21}\text{kg}\cdot \text{m/s}\end{array}$

Energy required:

  $\begin{array}{c}\Delta E=\sqrt{\Delta p^2{c}^2+m_o{}^2{c}^4}\\ =\sqrt{{\left(5.28\times {10}^{-21}\right)}^2{\left(3\times {10}^8\right)}^2+{\left(9.1\times {10}^{-31}\right)}^2{\left(3\times {10}^8\right)}^4}\\ =1.586\times {10}^{-12}\text{J}\\ \text{= 9}\text{.9 MeV}\end{array}$