Question

Suppose the velocity of an electron in an atom is known to an accuracy of 2.0 x 10³ (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1 nm size of the atom?

Short answer

The electron’s minimum uncertainty in velocity is: 2.9 x 10^-8m. Also, the value is 290 larger than size of the atom.

Step-by-step solution

Determine the formulas:

Determine the uncertainty principle formula:

$\mathrm{\Delta }x\mathrm{\Delta }p=\frac{h}{2\pi }$

Here,$\mathrm{\Delta }x$is the uncertainty in the position and$\mathrm{\Delta }p$is the uncertainty in the momentum.

Consider the expression for the momentum is$$\mathrm{\Delta }p=m\mathrm{\Delta }v$$.

Determine the electron’s minimum uncertainty in velocity

Determine the uncertainty in position as:

$$\begin{array}{c}\mathrm{\Delta }x=\frac{6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}}{4\pi \times 9.11\times {10}^{-31}\mathrm{k}\mathrm{g}\times 2.0\times {10}^3\mathrm{m}{\mathrm{s}}^{-1}}\\ =2.9\times {10}^{-8}\mathrm{m}\end{array}$$

Compare the uncertainty with the size of the atom as follows:

$$\begin{array}{c}\frac{\mathrm{\Delta }x}{s}=\frac{2.9\times {10}^{-8}\mathrm{m}}{0.1\times {10}^{-9}}\\ =290\end{array}$$

Therefore, the uncertainty in position is 2.9 x 10^-8m. Also, the value is 290 larger than size of the atom.