Question

fill in all the empty boxes. an atom of carbon has a radius of 67. pm and the average orbital speed of the electrons in it is about 1.3 × 10⁷ m/s. calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of carbon. write your answer as a percentage of the average speed, and round it to 2 significant digits. % correct answer 3.3%

Explanation:

Step1: Recall Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle for position (\(\Delta x\)) and momentum (\(\Delta p\)) is \(\Delta x \Delta p \geq \frac{h}{4\pi}\), where \(h = 6.626\times 10^{-34}\, \text{J·s}\) and momentum \(p = mv\), so \(\Delta p = m\Delta v\) (mass of electron \(m = 9.109\times 10^{-31}\, \text{kg}\)). First, find \(\Delta x\): the uncertainty in position for an electron in an atom is approximately twice the atomic radius (since the electron could be anywhere within the atom, so \(\Delta x = 2\times 67\, \text{pm}= 134\, \text{pm}= 134\times 10^{-12}\, \text{m}\)).

Step2: Solve for \(\Delta v\)

From \(\Delta x \cdot m\Delta v \geq \frac{h}{4\pi}\), we can solve for \(\Delta v\):
\(\Delta v \geq \frac{h}{4\pi m \Delta x}\)
Substitute values:
\(h = 6.626\times 10^{-34}\, \text{J·s}\), \(m = 9.109\times 10^{-31}\, \text{kg}\), \(\Delta x = 134\times 10^{-12}\, \text{m}\)
\(\Delta v = \frac{6.626\times 10^{-34}}{4\pi\times 9.109\times 10^{-31}\times 134\times 10^{-12}}\)
Calculate denominator: \(4\pi\times 9.109\times 10^{-31}\times 134\times 10^{-12} \approx 4\times 3.1416\times 9.109\times 10^{-31}\times 134\times 10^{-12} \approx 1.53\times 10^{-39}\)
\(\Delta v \approx \frac{6.626\times 10^{-34}}{1.53\times 10^{-39}} \approx 4.33\times 10^{5}\, \text{m/s}\)

Step3: Calculate percentage uncertainty

Average speed \(v = 1.3\times 10^{7}\, \text{m/s}\)
Percentage uncertainty \(= \frac{\Delta v}{v} \times 100 = \frac{4.33\times 10^{5}}{1.3\times 10^{7}} \times 100\)
\(\frac{4.33\times 10^{5}}{1.3\times 10^{7}} = \frac{4.33}{130} \approx 0.0333\)
Multiply by 100: \(0.0333\times 100 = 3.33\%\), round to 2 significant digits: \(3.3\%\)

Answer:

\(3.3\%\)