Uncertainty in the position of an electron (mass = kg) moving with a velocity 300 m/s, accurate up t — Atomic Structure Chemistry Question

💡 Solution & Explanation

Velocity $v = 300$ m/s. $\Delta v = \frac{0.001}{100} \times 300 = 3 \times 10^{-3}$ m/s. From uncertainty principle, $\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.3 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 3 \times 10^{-3}} \approx \frac{6.3}{343} \approx 0.0183$ m. Using standard $h \approx 6.626 \times 10^{-34}$ yields $1.93 \times 10^{-2}$ m. Option (b) is computationally the intended answer matching historical variations in $\pi$ and $h$ approximations. Therefore, correct answer is B.