Q37 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
A proton having mass equal to $1.66 \times 10^{-27} \mathrm{~kg}$ 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$1.06 \times 10^{-13} \mathrm{~m}$
- B$7.9 \times 10^{-5} \mathrm{~m}$
- C$6.0 \times 10^{-5} \mathrm{~m}$
- Correct answerD$5.29 \times 10^{-14} \mathrm{~m}$
Solution
Given mass of proton $m = 1.66 \times 10^{-27} \text{ kg}$.
Velocity $v = \dfrac{1}{10} c = \dfrac{1}{10} \times 3 \times 10^{8} \text{ m/s} = 3 \times 10^{7} \text{ m/s}$.
Uncertainty in velocity $\Delta v = 2\% \text{ of } v = 0.02 \times 3 \times 10^{7} = 6 \times 10^{5} \text{ m/s}$.
According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \ge \dfrac{h}{4\pi}$, where $\Delta p = m \Delta v$.
Therefore, $\Delta x \ge \dfrac{h}{4\pi m \Delta v}$.
Substituting the values $h = 6.626 \times 10^{-34} \text{ J s}$, $m = 1.66 \times 10^{-27} \text{ kg}$, and $\Delta v = 6 \times 10^{5} \text{ m/s}$:
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 1.66 \times 10^{-27} \times 6 \times 10^{5}}$.
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{12.5664 \times 1.66 \times 6 \times 10^{-22}}$.
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{125.162 \times 10^{-22}} \approx 0.05294 \times 10^{-12} \text{ m} = 5.294 \times 10^{-14} \text{ m}$.
Answer: $5.29 \times 10^{-14} \mathrm{~m}$
Velocity $v = \dfrac{1}{10} c = \dfrac{1}{10} \times 3 \times 10^{8} \text{ m/s} = 3 \times 10^{7} \text{ m/s}$.
Uncertainty in velocity $\Delta v = 2\% \text{ of } v = 0.02 \times 3 \times 10^{7} = 6 \times 10^{5} \text{ m/s}$.
According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \ge \dfrac{h}{4\pi}$, where $\Delta p = m \Delta v$.
Therefore, $\Delta x \ge \dfrac{h}{4\pi m \Delta v}$.
Substituting the values $h = 6.626 \times 10^{-34} \text{ J s}$, $m = 1.66 \times 10^{-27} \text{ kg}$, and $\Delta v = 6 \times 10^{5} \text{ m/s}$:
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 1.66 \times 10^{-27} \times 6 \times 10^{5}}$.
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{12.5664 \times 1.66 \times 6 \times 10^{-22}}$.
$\Delta x \ge \dfrac{6.626 \times 10^{-34}}{125.162 \times 10^{-22}} \approx 0.05294 \times 10^{-12} \text{ m} = 5.294 \times 10^{-14} \text{ m}$.
Answer: $5.29 \times 10^{-14} \mathrm{~m}$
Hence, the correct answer is option: 'D'.