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 `5.0 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. `80 kg ms^(-1)`
C. `80xx10^(-26) kg ms^(-1)`
D. `5.0 xx 10^(-26) kg ms^(-1)`

Explanation

To solve the problem, we will use Heisenberg's Uncertainty Principle, which states that the product of the uncertainties in position (Δx) and momentum (Δp) is greater than or equal to a constant, specifically:

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

where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \).

### Step 1: Identify the given values
- The uncertainty in position (Δx) is given as \( 1.0 \, \text{nm} = 1.0 \times 10^{-9} \, \text{m} \).
- The uncertainty in momentum of the electron (Δp_e) is given as \( 5.0 \times 10^{-26} \, \text{kg m/s} \).

### Step 2: Calculate the minimum uncertainty in momentum (Δp) using Heisenberg's principle
We can rearrange the uncertainty principle to find the uncertainty in momentum:

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

Substituting the values:

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

### Step 3: Calculate the value
Calculating the denominator:

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

Now substituting this 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 the uncertainties
Since the uncertainty in momentum of the helium atom is not provided directly, we can assume that if the position uncertainty is the same for both the electron and the helium atom, the uncertainty in momentum for the helium atom (Δp_He) should also be similar to that of the electron (Δp_e).

Thus, we can conclude that the minimum uncertainty in the momentum of the helium atom is:

\[ \Delta p_{He} \approx 5.0 \times 10^{-26} \, \text{kg m/s} \]

### Final Answer
The minimum uncertainty in the measurement of the momentum of the helium atom is \( 5.0 \times 10^{-26} \, \text{kg m/s} \).

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