Answer
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Work Step by Step
The Heisenberg uncertainty principle states that \( \Delta x \cdot \Delta p \geq \frac{\hbar}{4\pi} \), where \( \Delta x \) is the uncertainty in position, \( \Delta p \) is the uncertainty in momentum, and \( \hbar \) is the reduced Planck constant.
a. For an electron with \( \Delta v = 0.100 \, \text{m/s} \), we can calculate \( \Delta x \) using the mass and velocity of the electron. Given that the mass of an electron is approximately \( 9.109 \times 10^{-31} \, \text{kg} \), we have:
\[
\Delta p = m \cdot \Delta v = (9.109 \times 10^{-31} \, \text{kg}) \cdot (0.100 \, \text{m/s}) = 9.109 \times 10^{-32} \, \text{kg m/s}
\]
Now, rearranging the uncertainty principle equation to solve for \( \Delta x \), we get:
\[
\Delta x \geq \frac{\hbar}{4\pi \cdot \Delta p}
\]
Substituting the known values (\( \hbar = 6.626 \times 10^{-34} \, \text{J s} \)):
\[
\Delta x \geq \frac{(6.626 \times 10^{-34} \, \text{J s})}{4\pi \cdot (9.109 \times 10^{-32} \, \text{kg m/s})} \approx 5.8 \times 10^{-4} \, \text{m}
\]
The uncertainty in position for the electron is approximately \( 5.8 \times 10^{-4} \, \text{m} \). This is comparable to the size of a hydrogen atom, which is typically around \( 5 \times 10^{-4} \, \text{m} \).
b. For a baseball with mass \( 145 \, \text{g} \) and \( \Delta v = 0.100 \, \text{m/s} \), we calculate \( \Delta x \) using the same uncertainty principle equation. First, we need to convert the mass of the baseball to kilograms (\( 0.145 \, \text{kg} \)).
\[
\Delta p = m \cdot \Delta v = (0.145 \, \text{kg}) \cdot (0.100 \, \text{m/s}) = 0.0145 \, \text{kg m/s}
\]
Now, substituting into the uncertainty principle equation:
\[
\Delta x \geq \frac{(1.055 \times 10^{-34} \, \text{J s})}{2 \cdot (0.0145 \, \text{kg m/s})} \approx 3.63 \times 10^{-33} \, \text{m}
\]
The uncertainty in position for the baseball is extremely small, approximately \( 3.63 \times 10^{-33} \, \text{m} \), which is significantly smaller than the size of a baseball.
