Question

an atom of krypton has a radius (r_{kr}=88) pm and an average speed in the gas phase at (25^{circ}c) of 172 m/s. suppose the speed of a krypton atom at (25^{circ}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 in a multiple of (r_{kr}) 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 (r_{kr}), you would enter 42 (r_{kr}) as your answer.

Explanation:

Step1: Recall the Heisenberg uncertainty principle

$\Delta x\Delta p\geq\frac{h}{4\pi}$, where $\Delta x$ is the uncertainty in position, $\Delta p = m\Delta v$ is the uncertainty in momentum, $h = 6.626\times10^{- 34}\text{ J}\cdot\text{s}$ is Planck's constant, and $m$ is the mass of the atom. First, find the mass of a krypton - 88 atom. The molar mass of krypton - 88 is $M = 88\text{ g/mol}=0.088\text{ kg/mol}$. Using $m=\frac{M}{N_A}$, where $N_A = 6.022\times10^{23}\text{ mol}^{-1}$, we have $m=\frac{0.088}{6.022\times10^{23}}\text{ kg}\approx1.46\times10^{-25}\text{ kg}$.

Step2: Determine the uncertainty in velocity

The speed is measured within $0.010\%$. So the uncertainty in velocity $\Delta v = 0.0001\times172\text{ m/s}= 0.0172\text{ m/s}$.

Step3: Calculate the uncertainty in position

From $\Delta x\Delta p\geq\frac{h}{4\pi}$ and $\Delta p = m\Delta v$, we can solve for $\Delta x$. Rearranging gives $\Delta x=\frac{h}{4\pi m\Delta v}$. Substitute $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$, $m = 1.46\times10^{-25}\text{ kg}$, and $\Delta v=0.0172\text{ m/s}$ into the formula:
\[

$$\begin{align*} \Delta x&=\frac{6.626\times 10^{-34}}{4\pi\times1.46\times10^{-25}\times0.0172}\\ &=\frac{6.626\times 10^{-34}}{4\pi\times2.5112\times10^{-27}}\\ &=\frac{6.626\times 10^{-34}}{3.154\times10^{-26}}\\ &\approx2.1\times10^{-8}\text{ m} = 21\text{ nm} \end{align*}$$

\]
Since the answer should be in terms of $r_K$, and $r_K = 88\text{ pm}=88\times10^{-12}\text{ m}$, then $\frac{\Delta x}{r_K}=\frac{2.1\times10^{-8}}{88\times10^{-12}}\approx240$. Rounding to 2 significant figures, the answer is $24r_K$.

Answer:

$24r_K$