Using the Heisenberg uncertainty principle, calculate \(\Delta x\) for each of the following. a. an electron with \(\Delta v=0.100 \mathrm{~m} / \mathrm{s}\) b. a baseball (mass \(=145 \mathrm{~g}\) ) with \(\Delta v=0.100 \mathrm{~m} / \mathrm{s}\) c. How does the answer in part a compare with the size of a hydrogen atom? d. How does the answer in part b correspond to the size of a baseball?

a. For the electron, we have \(\Delta x_e \geq \dfrac{6.626 \times 10^{-34} J\cdot s}{4\pi \cdot (9.11 \times 10^{-31} kg) \cdot (0.100 \frac{m}{s})} \approx 1.83 \times 10^{-9}\: m\). b. For the baseball, we have \(\Delta x_b \geq \dfrac{6.626 \times 10^{-34} J\cdot s}{4\pi \cdot (0.145 kg) \cdot (0.100 \frac{m}{s})} \approx 3.01 \times 10^{-29}\: m\). c. The value of \(\Delta x_e \approx 1.83 \times 10^{-9}\: m\) is close to the size of a hydrogen atom (\(1 \times 10^{-10}\: m\)), which implies that the uncertainty principle is significant for microscopic particles like electrons. d. The value of \(\Delta x_b \approx 3.01 \times 10^{-29}\: m\) is much smaller than the size of a baseball (\(0.073\: m\) diameter), suggesting that the uncertainty principle has little impact on macroscopic objects like baseballs.