Question

A proton having mass equal to \(1.66 \times 10^{-27}\) kg is accelerated to one tenth of the velocity of light. If its velocity can be measured to an accuracy of \(\pm 2%\), what would be the uncertainty in its position?

• A. \(7.9 \times 10^{-5}\) m
• B. \(5.29 \times 10^{-14}\) m
• C. \(6.0 \times 10^{-5}\) m
• D. \(1.06 \times 10^{-13}\) m

Hint:

Heisenberg's uncertainty principle is a fundamental concept in quantum mechanics. It shows that the more accurately we measure the momentum (or velocity), the less accurately we can determine the position of a particle.

Answer 1

The Correct Option is B

To solve this problem, we need to use Heisenberg's uncertainty principle, which is given by the relation: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] where: - \(\Delta x\) is the uncertainty in position, - \(\Delta p\) is the uncertainty in momentum, - \(\hbar\) is the reduced Planck's constant, \( \hbar = 1.0545718 \times 10^{-34} \, \text{J s}\). The uncertainty in momentum \(\Delta p\) can be written as: \[ \Delta p = m \Delta v \] where: - \(m = 1.66 \times 10^{-27} \, \text{kg}\) (mass of the proton), - \(\Delta v = 2% \, \text{of the velocity of the proton}\). The velocity of the proton is given as \(v = \frac{c}{10}\), where \(c = 3.0 \times 10^8 \, \text{m/s}\). Therefore: \[ v = 3.0 \times 10^7 \, \text{m/s} \] Now, the uncertainty in velocity \(\Delta v\) is: \[ \Delta v = 0.02 \times v = 0.02 \times 3.0 \times 10^7 = 6.0 \times 10^5 \, \text{m/s} \] Now, calculate the uncertainty in momentum \(\Delta p\): \[ \Delta p = m \Delta v = (1.66 \times 10^{-27}) \times (6.0 \times 10^5) = 9.96 \times 10^{-22} \, \text{kg m/s} \] Using the uncertainty principle: \[ \Delta x \geq \frac{\hbar}{2 \Delta p} \] Substitute the known values: \[ \Delta x \geq \frac{1.0545718 \times 10^{-34}}{2 \times 9.96 \times 10^{-22}} = 5.29 \times 10^{-14} \, \text{m} \] Therefore, the uncertainty in the position is \(5.29 \times 10^{-14} \, \text{m}\).