Question

Question:The uncertainty in the position of an electron along an $\mathit{x}$axis

is given as $\mathbf{50}\mathbf{}\mathbf{pm}$, which is about equal to the radius of a hydrogen

atom. What is the least uncertainty in any simultaneous measurement

of the momentum component of this electron?

Short answer

The least uncertainty in the momentum component $p_x$of this electron is

$∆p_x=2.1\times {10}^{-24}\frac{\mathrm{kgm}}{\mathrm{s}}$

Step-by-step solution

Identifying the data given in the question.

The uncertainty in the position of the electron is

$∆x=50\mathrm{pm}=50\times {10}^{-12}\mathrm{m}$

Concept used to solve the question

Heisenberg’s Uncertainty Principle

According to Heisenberg’s Uncertainty Principle, the uncertainty in position and momentum are related as,

$\mathbf{∆}\mathit{x}\mathbf{∆}{\mathit{p}}_{\mathbf{x}}\mathbf{\ge }\frac{\mathbf{h}}{\mathbf{2}\mathbf{\pi }}$

Where;

$\mathbf{∆}\mathit{x}$ is uncertainty in position along the axis.

$\mathbf{∆}{\mathit{p}}_{\mathbf{x}}$is uncertainty in momentum along the axis.

$\mathit{h}$is plank’s constant.

Finding the least uncertainty in the momentum component px

According to Heisenberg’s Uncertainty Principle

$∆x∆p_x\ge \frac{h}{2\mathrm{\pi }}$

The least uncertainty possible when

$∆x∆p_x=\frac{h}{2\mathrm{\pi }}$

Therefore,

$∆p_x=\frac{h}{2\mathrm{\pi }∆\mathrm{x}}$

Substituting the values,

$\begin{array}{rcl}∆p_x& =& \frac{6.634\times {10}^{-34}J.s}{2\mathrm{\pi }\left(50\times {10}^{-12}\mathrm{m}\right)}\\ & =& 2.1\times {10}^{-24}kgm{s}^{-1}\end{array}$

Hence the least uncertainty in the momentum component $p_x$of this electron is

$\begin{array}{rcl}∆p_x& =& 2.1\times {10}^{-24}kgm{s}^{-1}\end{array}$