Question

What is the smallest one-dimensional box in which you can confine an electron if you want to know for certain that the electron's speed is no more than $10\mathrm{m}/\mathrm{s}$ ?

Short answer

The smallest one-dimensional box is $18.2\times {10}^{-6}\mathrm{m}.$

Step-by-step solution

part  (a) step 1: Given information 

Electron's velocity limit: $10\mathrm{m}/\mathrm{s}$

part (a) step 2: formulas for solutions

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

Whereas, $\Delta x$is the change in position.

$\Delta p$is change in momentum.

h is Plank's constant $\left(h=6.63\times {10}^{-34}\mathrm{Js}\right)$

part (b) step 3: calculation

The range of electron's velocity is $10\mathrm{m}/\mathrm{s}$so it can be varied from $-10\mathrm{m}/\mathrm{s}$too $+10\mathrm{m}/\mathrm{s}.$

So the change in velocity is $20\mathrm{m}/\mathrm{s}.$

$\Delta p$Change in momentum can be given as $\Delta p=m\Delta v$where; $\mathrm{m}$is the mass of electron

$$\begin{gathered} \Delta p=m\Delta v \\ \Delta p=9.1\times {10}^{-31}\mathrm{kg}\times 20\mathrm{m}/\mathrm{s} \\ \Delta p=1.82\times {10}^{-29} \end{gathered}$$

Here, $\Delta x$is the uncertainty in the position of an electron which is equal to the dimension of the box L

So by using the formula $\Delta x\Delta p=\frac{h}{2}$, we can determine the dimension of the box.

$$\begin{gathered} \Delta x\Delta p=\frac{h}{2} \\ \Delta x=\frac{h}{2\Delta p}=\frac{6.63\times {10}^{-3\mathrm{Js}}}{2\times 1.82\times {10}^{-29}}=18.2\times {10}^{-6}\mathrm{m} \end{gathered}$$

The dimension of the smallest box should be

$18.2\times {10}^{-6}\mathrm{m}.$