Question

The minimum values of uncertainties involved in the determination of both the position and velocity of a particle are respectively \(1\times10^{-10}\,m\) and \(1\times10^{-10}\,ms^{-1}\). Then, the mass (in kg) of the particle is

• A. \(5.270\times10^{-15}\)
• B. \(5.270\times10^{-20}\)
• C. \(5.270\times10^{-16}\)
• D. \(5.270\times10^{-10}\)
• E. \(5.270\times10^{-14}\)

Hint:

Minimum uncertainty uses equality in Heisenberg relation.

Answer 1

The Correct Option is C

Concept: Heisenberg uncertainty principle: \[ \Delta x \cdot \Delta p = \frac{h}{4\pi} \quad \text{and} \quad \Delta p = m \Delta v \]

Step 1: Substitute momentum uncertainty. \[ \Delta x \cdot m \Delta v = \frac{h}{4\pi} \]

Step 2: Solve for mass. \[ m = \frac{h}{4\pi \Delta x \Delta v} \]

Step 3: Substitute values. \[ h = 6.626\times10^{-34} \] \[ \Delta x = 10^{-10}, \quad \Delta v = 10^{-10} \] \[ m = \frac{6.626\times10^{-34}}{4\pi \times 10^{-20}} \]

Step 4: Simplify. \[ m = \frac{6.626}{12.56} \times 10^{-14} \] \[ \approx 0.527 \times 10^{-14} = 5.27 \times 10^{-16} \]

Step 5: Final answer. \[ \boxed{5.27\times10^{-16}\,kg} \]