Question

The position of both, an electron and a helium atom is known within 1.0 nm. Further the momentum of the electron is known within \( 5.0 \times 10^{-26} \, \text{kg} \, \text{ms}^{-1} \). The minimum uncertainty in the measurement of the momentum of the helium atom is:

• A. \( 50 \, \text{kg} \, \text{ms}^{-1} \)
• B. \( 80 \, \text{kg} \, \text{ms}^{-1} \)
• C. \( 8.0 \times 10^{-26} \, \text{kg} \, \text{ms}^{-1} \)
• D. \( 5.0 \times 10^{-26} \, \text{kg} \, \text{ms}^{-1} \)

Hint:

The uncertainty in momentum decreases with the mass of the particle. The helium atom's larger mass results in a smaller uncertainty compared to the electron.

Answer 1

The Correct Option is D

The uncertainty principle \( \Delta x \, \Delta p \geq \frac{h}{4\pi} \) relates the uncertainty in position and momentum. Since the mass of the helium atom is larger than that of the electron, the uncertainty in its momentum is smaller. Therefore, the minimum uncertainty in the momentum of the helium atom is \( 5.0 \times 10^{-26} \, \text{kg} \, \text{ms}^{-1} \).

Step 2: Conclusion.
Thus, the correct answer is \( 5.0 \times 10^{-26} \, \text{kg} \, \text{ms}^{-1} \).