Question

A fast moving particle of mass 6.63×10^-28g can be located with an accuracy of 1A. The uncertainty in its velocity (in ms^-1) is about (h=6.63×10^-34 Js)

• A. 8×10³
• B. 8×10⁴
• C. 8×10⁵
• D. 8×10⁶
• E. 8×10⁷

Answer 1

The Correct Option is C

This problem can be solved using Heisenberg's uncertainty principle, which states: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where \( \Delta x \) is the uncertainty in position and \( \Delta p \) is the uncertainty in momentum. Momentum \( p \) is given by: \[ p = mv \] where \( m \) is the mass and \( v \) is the velocity of the particle. The uncertainty in momentum \( \Delta p \) is: \[ \Delta p = m \Delta v \] where \( \Delta v \) is the uncertainty in velocity. Now, the uncertainty principle equation becomes: \[ \Delta x \cdot m \Delta v \geq \frac{h}{4\pi} \] Solving for \( \Delta v \): \[ \Delta v \geq \frac{h}{4\pi m \Delta x} \] Given: - \( h = 6.63 \times 10^{-34} \) J·s, - \( m = 6.63 \times 10^{-28} \) g \( = 6.63 \times 10^{-31} \) kg, - \( \Delta x = 1 \) Å \( = 10^{-10} \) m, Substitute the values into the equation: \[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4\pi \times 6.63 \times 10^{-31} \times 10^{-10}} = 8 \times 10^5 \, \text{ms}^{-1} \]

Thus, the correct option is (C) : \(8×10^5\)

Answer 2

To find the uncertainty in velocity, we use the Heisenberg Uncertainty Principle: 

\(\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\)
Since \(\Delta p = m \cdot \Delta v\), we have: 

\[ \Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi} \] \[ \Delta v \geq \frac{h}{4\pi m \Delta x} \]

 Given:
- \(h = 6.63 \times 10^{-34}\, \text{Js}\)
- \(m = 6.63 \times 10^{-28}\, \text{g} = 6.63 \times 10^{-31}\, \text{kg}\)
- \(\Delta x = 1\, \text{Å} = 1 \times 10^{-10}\, \text{m}\) 
Now substitute the values: \[ \Delta v \geq \frac{6.63 \times 10^{-34}}{4\pi \times 6.63 \times 10^{-31} \times 1 \times 10^{-10}} \] \[ \Delta v \geq \frac{1}{4\pi \times 10^{-41}} \approx \frac{1}{1.26 \times 10^{-40}} \approx 7.96 \times 10^5 \] 
Final Answer: 8×10⁵ ms^-1