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NEET CHEMISTRYMedium

Given the following data: Mass of electron: 9.11 × 10⁻³¹ kg Planck's constant: 6.626 × 10⁻³⁴ Js In light of the data given above, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å will be:

A

5.79 × 10⁶ ms⁻¹

B

5.79 × 10⁷ ms⁻¹

C

5.79 × 10⁸ ms⁻¹

D

5.79 × 10⁵ ms⁻¹

Step-by-Step Solution

According to the Heisenberg Uncertainty Principle, the product of the uncertainty in position (Δx\Delta x) and the uncertainty in momentum (Δp\Delta p) is given by: Δx⋅Δp≥h4π\Delta x \cdot \Delta p \ge \frac{h}{4\pi} Since Δp=m⋅Δv\Delta p = m \cdot \Delta v, the equation becomes: Δx⋅(m⋅Δv)≥h4π  ⟹  Δv≥h4π⋅m⋅Δx\Delta x \cdot (m \cdot \Delta v) \ge \frac{h}{4\pi} \implies \Delta v \ge \frac{h}{4\pi \cdot m \cdot \Delta x}

  1. Convert Units:
  • Distance (Δx\Delta x) = 0.1 A˚=0.1×10−10 m=10−11 m0.1 \text{ \AA} = 0.1 \times 10^{-10} \text{ m} = 10^{-11} \text{ m} .
  1. Substitute Values: h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \text{ Js} m=9.11×10−31 kgm = 9.11 \times 10^{-31} \text{ kg}
  • π≈3.14159\pi \approx 3.14159
  1. Calculation: Δv=6.626×10−344×3.14159×(9.11×10−31)×10−11\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14159 \times (9.11 \times 10^{-31}) \times 10^{-11}} Δv=6.626×10−34114.48×10−42\Delta v = \frac{6.626 \times 10^{-34}}{114.48 \times 10^{-42}} Δv=0.05788×108 ms−1\Delta v = 0.05788 \times 10^8 \text{ ms}^{-1} Δv=5.79×106 ms−1\Delta v = 5.79 \times 10^6 \text{ ms}^{-1}
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