Question

an atom of lead has a radius of 154 pm and the average orbital speed of the electrons in it is about 1.8×10^8 m/s. calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. write your answer as a percentage of the average speed, and round it to 2 significant digits.

Explanation:

Step1: Recall the Heisenberg uncertainty - principle formula

The Heisenberg uncertainty - principle is given by $\Delta x\Delta p\geq\frac{h}{4\pi}$, where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h = 6.63\times10^{-34}\ J\cdot s$ is Planck's constant. For an electron in an atom, we can assume the uncertainty in position $\Delta x$ is of the order of the atomic radius. Here, $r = 154\ pm=154\times10^{-12}\ m$, so $\Delta x\approx r = 154\times10^{-12}\ m$.

Step2: Express momentum uncertainty in terms of velocity uncertainty

Momentum $p = mv$, and the uncertainty in momentum $\Delta p=m\Delta v$, where $m = 9.11\times10^{-31}\ kg$ is the mass of an electron. From the Heisenberg uncertainty - principle $\Delta p\geq\frac{h}{4\pi\Delta x}$.

Step3: Calculate the minimum uncertainty in velocity

$\Delta v\geq\frac{h}{4\pi m\Delta x}$. Substitute $h = 6.63\times10^{-34}\ J\cdot s$, $m = 9.11\times10^{-31}\ kg$, and $\Delta x = 154\times10^{-12}\ m$ into the formula:
\[

$$\begin{align*} \Delta v&\geq\frac{6.63\times10^{-34}}{4\pi\times9.11\times10^{-31}\times154\times10^{-12}}\\ &=\frac{6.63\times10^{-34}}{4\pi\times9.11\times154\times10^{-43}}\\ &=\frac{6.63\times10^{-34}}{4\pi\times1.403\times10^{-38}}\\ &\approx3.7\times10^{4}\ m/s \end{align*}$$

\]

Step4: Calculate the percentage of the average orbital speed

The average orbital speed $v_{avg}=1.8\times10^{8}\ m/s$. The percentage is $\frac{\Delta v}{v_{avg}}\times100\%$. Substitute $\Delta v = 3.7\times10^{4}\ m/s$ and $v_{avg}=1.8\times10^{8}\ m/s$:
\[

$$\begin{align*} \frac{\Delta v}{v_{avg}}\times100\%&=\frac{3.7\times10^{4}}{1.8\times10^{8}}\times100\%\\ &=\frac{3.7\times10^{4}\times100}{1.8\times10^{8}}\\ &=\frac{3.7\times10^{6}}{1.8\times10^{8}}\\ &\approx0.021\% \end{align*}$$

\]

Answer:

$0.021\%$