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Chemistry · Structure of atom
हिन्दीAn electron is moving with a velocity of \( 3 \times 10^6 \text{ m s}^{-1} \). If the uncertainty in its velocity is \( 0.01\% \), calculate the uncertainty in its position (\( \Delta x \). Given: mass of electron \( m = 9.11 \times 10^{-31} \text{ kg} \), \( h = 6.626 \times 10^{-34} \text{ J s} \).
Correct answer · Option 3
\( 1.93 \times 10^{-4} \text{ m} \)
Explanation
Using Heisenberg's uncertainty relation \( \Delta x \cdot \Delta v_x \ge \frac{h}{4\pi m} \), where \( \Delta v_x = \frac{0.01}{100} \times 3 \times 10^6 = 300 \text{ m s}^{-1} \). Substituting values gives \( \Delta x = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 9.11 \times 10^{-31} \times 300} \approx 1.93 \times 10^{-4} \text{ m} \).
