Question

an atom of argon has a radius ( r_{\text{ar}} = 71. , \text{pm} ) and an average speed in the gas phase at ( 25^circ \text{c} ) of ( 249. , \text{m/s} ). suppose the speed of an argon atom at ( 25^circ \text{c} ) has been measured to within ( 0.010% ). calculate the smallest possible length of box inside of which the atom could be known to be located with certainty. write your answer as a multiple of ( r_{\text{ar}} ) and round it to 2 significant figures. for example, if the smallest box the atom could be in turns out to be 42.0 times the radius of an atom of argon, you would enter \42.0 ( r_{\text{ar}} )\ as your answer.

Explanation:

Step1: Recall Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle is given by \(\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\), where \(\Delta x\) is the uncertainty in position, \(\Delta p = m\Delta v\) is the uncertainty in momentum, \(m\) is the mass of the particle, \(\Delta v\) is the uncertainty in velocity, and \(h = 6.626\times 10^{-34}\, \text{J}\cdot\text{s}\) is Planck's constant.

First, find the uncertainty in velocity \(\Delta v\). The average speed \(v = 249\, \text{m/s}\), and the uncertainty is \(0.010\%\) of \(v\). So \(\Delta v= 0.00010\times v=0.00010\times249\, \text{m/s}= 0.0249\, \text{m/s}\).

Step2: Calculate mass of argon atom

The molar mass of argon \(M = 39.95\, \text{g/mol}=39.95\times 10^{-3}\, \text{kg/mol}\). Using Avogadro's number \(N_A = 6.022\times 10^{23}\, \text{mol}^{-1}\), the mass of one argon atom \(m=\frac{M}{N_A}=\frac{39.95\times 10^{-3}\, \text{kg/mol}}{6.022\times 10^{23}\, \text{mol}^{-1}}\approx 6.634\times 10^{-26}\, \text{kg}\).

Step3: Calculate uncertainty in momentum \(\Delta p\)

\(\Delta p = m\Delta v\). Substituting the values of \(m\) and \(\Delta v\): \(\Delta p=(6.634\times 10^{-26}\, \text{kg})\times(0.0249\, \text{m/s})\approx 1.652\times 10^{-27}\, \text{kg}\cdot\text{m/s}\).

Step4: Calculate uncertainty in position \(\Delta x\)

From the Heisenberg Uncertainty Principle \(\Delta x \geq \frac{h}{4\pi\Delta p}\). Substituting \(h = 6.626\times 10^{-34}\, \text{J}\cdot\text{s}\) and \(\Delta p = 1.652\times 10^{-27}\, \text{kg}\cdot\text{m/s}\):

\[ \Delta x\geq\frac{6.626\times 10^{-34}}{4\pi\times1.652\times 10^{-27}} \]

\[ \Delta x\geq\frac{6.626\times 10^{-34}}{2.073\times 10^{-26}}\approx 3.196\times 10^{-8}\, \text{m} \]

Step5: Convert \(\Delta x\) to terms of \(r_{\text{Ar}}\)

The radius \(r_{\text{Ar}} = 71\, \text{pm}=71\times 10^{-12}\, \text{m}\). Let the length of the box be \(L = \Delta x\) (since the smallest box length is the uncertainty in position). We need to find \(n\) such that \(L = n\times r_{\text{Ar}}\).

\[ n=\frac{\Delta x}{r_{\text{Ar}}}=\frac{3.196\times 10^{-8}\, \text{m}}{71\times 10^{-12}\, \text{m}} \]

\[ n=\frac{3.196\times 10^{4}}{71}\approx 450.14 \]

Rounding to 2 significant figures, \(n\approx 4.5\times 10^{2}\) or \(4.5\times 10^{2}r_{\text{Ar}}\) (Wait, let's check the calculation again. Wait, maybe I made a mistake in the order of magnitude. Wait, \(r_{\text{Ar}} = 71\,\text{pm}=71\times 10^{-12}\, \text{m}\), \(\Delta x = 3.196\times 10^{-8}\, \text{m}=31960\times 10^{-12}\, \text{m}\). Then \(n=\frac{31960\times 10^{-12}}{71\times 10^{-12}}=\frac{31960}{71}\approx 450\), which is \(4.5\times 10^{2}\) when rounded to two significant figures. Wait, but let's re - check the Heisenberg formula. Wait, the Heisenberg Uncertainty Principle is \(\Delta x \Delta p\geq\frac{h}{4\pi}\), so \(\Delta x\geq\frac{h}{4\pi\Delta p}\). Let's recalculate \(\Delta p\):

\(m=\frac{39.95\times 10^{-3}\, \text{kg/mol}}{6.022\times 10^{23}\, \text{mol}^{-1}} = 6.634\times 10^{-26}\, \text{kg}\) (correct). \(\Delta v = 0.010\% \text{ of }249\, \text{m/s}=0.0001\times249 = 0.0249\, \text{m/s}\) (correct). \(\Delta p=m\Delta v=6.634\times 10^{-26}\times0.0249 = 1.652\times 10^{-27}\, \text{kg}\cdot\text{m/s}\) (correct).

\(h = 6.626\times 10^{-34}\, \text{J}\cdot\text{s}\), so \(\frac{h}{4\pi\Delta p}=\frac{6.626\times 10^{-34}}{4\times3.1416\times1.652\times 10^{-27}}\)

\(4\times3.1416\times1.652\times 10^{-27}\approx 2.073\times 10^{-26}\)

\(\frac{6.626\times 10^{-34}}{2.073\times 10^{-26}}\approx 3.196\times 10^{-8}\, \text{m}\) (correct).

\(r_{\text{Ar}} = 71\, \text{pm}=71\times 10^{-12}\, \text{m}\), so \(\frac{\Delta x}{r_{\text{Ar}}}=\frac{3.196\times 10^{-8}}{71\times 10^{-12}}=\frac{3.196\times 10^{4}}{71}\approx 450\), which is \(4.5\times 10^{2}\) when rounded to two significant figures. Wait, but maybe I messed up the significant figures. Wait, the given values: \(r_{\text{Ar}} = 71\, \text{pm}\) (two significant figures), speed \(249\, \text{m/s}\) (three), uncertainty \(0.010\%\) (two significant figures). So the result should have two significant figures. So \(n\approx 4.5\times 10^{2}\) or \(4.5\times 10^{2}r_{\text{Ar}}\)? Wait, no, \(450\) with two significant figures is \(4.5\times 10^{2}\), but let's check the calculation again. Wait, \(3.196\times 10^{-8}\, \text{m}\) divided by \(71\times 10^{-12}\, \text{m}\):

\(3.196\times 10^{-8}\div(71\times 10^{-12})=(3.196\div71)\times 10^{4}\approx0.04501\times 10^{4}=450.1\), which is \(4.5\times 10^{2}\) when rounded to two significant figures. So the smallest possible length of the box is \(4.5\times 10^{2}r_{\text{Ar}}\) or \(4.5\times 10^{2}r_{\text{Ar}}\) (wait, maybe I made a mistake in the Heisenberg formula application. Wait, the box length is the uncertainty in position, and we need to express it as a multiple of \(r_{\text{Ar}}\). So the calculation seems correct.

Answer:

\(4.5\times 10^{2}r_{\text{Ar}}\) (or \(450r_{\text{Ar}}\) when considering two significant figures, but \(4.5\times 10^{2}r_{\text{Ar}}\) is more appropriate for two significant figures)