A bullet with mass \(10.000 \mathrm{g}\) has a speed of $300.00 \mathrm{m} / \mathrm{s}\( the speed is accurate to within \)0.04 \% .$ (a) Estimate the minimum uncertainty in the position of the bullet, according to the uncertainty principle. (b) An electron has a speed of $300.00 \mathrm{m} / \mathrm{s}\(, accurate to \)0.04 \% .$ Estimate the minimum uncertainty in the position of the electron. (c) What can you conclude from these results?

The Heisenberg Uncertainty Principle formula is given by: \(\Delta x \Delta p \geq \dfrac{\hbar}{2}\) where \(\Delta x\) is the uncertainty in position, \(\Delta p\) is the uncertainty in momentum, and \(\hbar\) is the reduced Planck's constant, approximately equal to \(1.055 \times 10^{-34} \mathrm{Js}\).

For both the bullet and the electron, the given speed uncertainty is \(0.04 \% = 0.0004\). (a) Bullet: Mass of the bullet is \(10.000 \mathrm{g} = 0.01\mathrm{kg}\) Speed of the bullet is \(300.00 \mathrm{m/s}\) Therefore, the momentum of the bullet is: \(p_{bullet} = m_{bullet} \times v_{bullet} = 0.01\mathrm{kg} \times 300.00 \mathrm{m/s} = 3\mathrm{kgm/s}\). Momentum uncertainty of the bullet \(\Delta p_{bullet} = p_{bullet} \times speed\_uncertainty = 3\mathrm{kgm/s} \times 0.0004 = 0.0012\mathrm{kgm/s}\) (b) Electron: Mass of the electron is approximately \(9.11 \times 10^{-31}\mathrm{kg}\) Speed of the electron is \(300.00 \mathrm{m}s\) Therefore, the momentum of the electron is: \(p_{electron} = m_{electron} \times v_{electron} = 9.11 \times 10^{-31}\mathrm{kg} \times 300.00 \mathrm{m/s} = 2.73 \times 10^{-28}\mathrm{kgm/s}\). Momentum uncertainty of the electron \(\Delta p_{electron} = p_{electron} \times speed\_uncertainty = 2.73 \times 10^{-28}\mathrm{kgm/s} \times 0.0004 = 1.092 \times 10^{-31}\mathrm{kgm/s}\)

Now, we'll use the Heisenberg Uncertainty Principle formula to find the minimum position uncertainty for both the bullet and the electron. (a) Bullet: \(\Delta x_{bullet} \geq \dfrac{\hbar}{2\Delta p_{bullet}} \geq \dfrac{1.055 \times 10^{-34}\mathrm{Js}}{2\times 0.0012\mathrm{kgm/s}} \geq 4.40 \times 10^{-32}\mathrm{m}\) (b) Electron: \(\Delta x_{electron} \geq \dfrac{\hbar}{2\Delta p_{electron}} \geq \dfrac{1.055 \times 10^{-34}\mathrm{Js}}{2\times 1.092 \times 10^{-31}\mathrm{kgm/s}} \geq 4.83 \times 10^{-4}\mathrm{m}\)

From the results, we can conclude the following: (a) The minimum uncertainty in the position of the bullet is extremely small, approximately \(4.40 \times 10^{-32} \mathrm{m}\) which is negligible in practical situations. (b) The minimum uncertainty in the position of the electron is much larger, approximately \(4.83 \times 10^{-4} \mathrm{m}\). The results indicate that the uncertainty principle has a more significant effect on small particles like electrons, and a negligible effect on macroscopic objects like bullets.