The position of both an electron and a helium atom is known within `1.0 nm` and the momentum of the electron is known within `50 xx 10^-26 kg ms^-1`. The minimum uncertainty in the measurement of the momentum of the helium atom is.

Options

A. `50 kg ms^-1`
B. `60 kg ms^-1`
C. `80 xx 10^-26 kg ms^-1`
D. `50 xx 10^-26 kg ms^-1`

Explanation

To find the minimum uncertainty in the measurement of the momentum of the helium atom, we can use the Heisenberg Uncertainty Principle, which states that the uncertainty in position (Δx) and the uncertainty in momentum (Δp) are related by the equation:

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

Where:
- \( \Delta x \) is the uncertainty in position,
- \( \Delta p \) is the uncertainty in momentum,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)).

### Step 1: Identify the given values
- The uncertainty in position for both the electron and the helium atom is given as \( \Delta x = 1.0 \, \text{nm} = 1.0 \times 10^{-9} \, \text{m} \).
- The uncertainty in momentum for the electron is given as \( \Delta p_e = 50 \times 10^{-26} \, \text{kg m/s} \).

### Step 2: Use the Heisenberg Uncertainty Principle
According to the Heisenberg Uncertainty Principle, we can express the relationship between the uncertainties in position and momentum:

\[ \Delta p \geq \frac{h}{4\pi \Delta x} \]

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:

\[ \Delta p \geq \frac{6.626 \times 10^{-34} \, \text{Js}}{4\pi (1.0 \times 10^{-9} \, \text{m})} \]

Calculating the denominator:

\[ 4\pi (1.0 \times 10^{-9}) \approx 1.25664 \times 10^{-9} \, \text{m} \]

Now substituting this value back into the equation:

\[ \Delta p \geq \frac{6.626 \times 10^{-34}}{1.25664 \times 10^{-9}} \approx 5.28 \times 10^{-25} \, \text{kg m/s} \]

### Step 4: Compare with the uncertainty in momentum of the electron
Since the uncertainties in position for both the electron and helium atom are the same, we can conclude that the uncertainty in momentum for the helium atom will also be the same as that of the electron.

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

\[ \Delta p_{He} = 50 \times 10^{-26} \, \text{kg m/s} \]