Question

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

• A. $7.0 \times 10^{-26} kg\, ms ^{-1}$
• B. $5.0 \times 10^{-26} kg \,ms ^{-1}$
• C. $8 \cdot 0 \times 10^{-26} \,kg\, ms ^{-1}$
• D. $6 \cdot 0 \times 10^{-26} kg \,ms ^{-1}$

Answer 1

The Correct Option is B

To find the minimum uncertainty in the measurement of the momentum of the helium atom, we will use the Heisenberg Uncertainty Principle. The principle states that the uncertainty in position (\(\Delta x\)) and the uncertainty in momentum (\(\Delta p\)) are related by:

\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

Where h is Planck's constant, approximately 6.626 \times 10^{-34} \, Js.

Given, for both the electron and helium atom:

• Uncertainty in position, \Delta x = 1.0 \, nm = 1.0 \times 10^{-9} \, m

First, let's calculate the minimum uncertainty in the momentum (\(\Delta p\)):

\Delta p \geq \frac{6.626 \times 10^{-34}}{4\pi \times 1.0 \times 10^{-9}}

Calculating the right side:

\Delta p \geq \frac{6.626 \times 10^{-34}}{12.5664 \times 10^{-9}}

\Delta p \geq 5.274 \times 10^{-26} \, kg \, ms^{-1}

This calculated minimum uncertainty in momentum is approximately 5.0 \times 10^{-26} \, kg \, ms^{-1}, which is already given as the momentum uncertainty for the electron.

Since the uncertainty in the position (\(\Delta x\)) is the same for both the electron and helium atom, the minimum uncertainty in the momentum of the helium atom remains the same as for the electron, according to the Heisenberg Uncertainty Principle.

Thus, the minimum uncertainty in the measurement of the momentum of the helium atom is:

5.0 \times 10^{-26} \, kg \, ms^{-1}

The correct answer is 5.0 \times 10^{-26} kg \, ms^{-1}.