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 `5.0xx10^(-26) kgms^(-1)` The minimum uncertainty in the measurement of the momentum of the helium atom is

Options

A. `6.0 xx10^(-26) kg ms ^(-1)`
B. `60 kgms^(-1)`
C. `50 kgms ^(-1)`
D. `5.0 xx10^(-26) kgms^(-1)`

Answer

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

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

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

### Step-by-Step Solution:

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 \( \Delta p_e = 5.0 \times 10^{-26} \, \text{kg m/s} \).

2. **Apply the Heisenberg Uncertainty Principle for the electron**:
\[ \Delta x_e \cdot \Delta p_e \geq \frac{h}{4\pi} \]
Substituting the known values:
\[ (1.0 \times 10^{-9} \, \text{m}) \cdot (5.0 \times 10^{-26} \, \text{kg m/s}) \geq \frac{6.626 \times 10^{-34}}{4\pi} \]

3. **Calculate the right side of the inequality**:
\[ \frac{6.626 \times 10^{-34}}{4\pi} \approx \frac{6.626 \times 10^{-34}}{12.566} \approx 5.283 \times 10^{-35} \, \text{kg m/s} \]

4. **Rearranging the equation for helium**:
Since the uncertainty in position for the helium atom is the same as that for the electron:
\[ \Delta x_h = \Delta x_e = 1.0 \times 10^{-9} \, \text{m} \]
We can write:
\[ \Delta x_h \cdot \Delta p_h \geq \frac{h}{4\pi} \]
Therefore:
\[ (1.0 \times 10^{-9} \, \text{m}) \cdot \Delta p_h \geq 5.283 \times 10^{-35} \, \text{kg m/s} \]

5. **Solve for Δp_h**:
\[ \Delta p_h \geq \frac{5.283 \times 10^{-35}}{1.0 \times 10^{-9}} = 5.283 \times 10^{-26} \, \text{kg m/s} \]

6. **Comparing with the uncertainty of the electron**:
Since the uncertainty in momentum of the electron is \( 5.0 \times 10^{-26} \, \text{kg m/s} \), we can conclude that the minimum uncertainty in the measurement of the momentum of the helium atom is the same:
\[ \Delta p_h = 5.0 \times 10^{-26} \, \text{kg m/s} \]

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