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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?

Short Answer

Expert verified
Answer: The position uncertainty for the bullet is approximately \(4.40 \times 10^{-32} \mathrm{m}\), while for the electron, it is approximately \(4.83 \times 10^{-4} \mathrm{m}\). The position uncertainty of the electron is much larger than that of the bullet, demonstrating the more significant effect of the uncertainty principle on small particles like electrons compared to macroscopic objects like bullets.

Step by step solution

01

(Step 1: Write the Heisenberg Uncertainty Principle formula)

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 \(1.055 \times 10^{-34} \mathrm{Js}\).
02

(Step 2: Calculate the momentum uncertainty of the bullet and the electron)

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 \(\Delta p_{electron} = p_{electron} \times speed\_uncertainty = 2.73 \times 10^{-28}\mathrm{kgm/s} \times 0.0004 = 1.092 \times 10^{-31}\mathrm{kgm/s}\)
03

(Step 3: Calculate the position uncertainty of the bullet and the electron)

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: \(\Delta x_{electron} \geq \dfrac{\hbar}{2\Delta p_{electron}} \geq \dfrac{1.055 \times 10^{-34}\mathrm{Js}}{2\times 1.092 \times 10^{-31}\mathrm{kgm/s}} \geq 4.83 \times 10^{-4}\mathrm{m}\)
04

(Step 4: Conclusion)

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.

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