Uncertainity in position of a `e^(-)` and He is similar. If uncertainity in momentum of `e^(-)` is `32xx10^(5)`, then uncertainity in momentum of He will be :

Options

A. `32xx10^(5)`
B. `16xx10^(5)`
C. `8xx10^(5)`
D. None of these

Answer

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

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

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

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Uncertainty in momentum of the electron (Δp_e) = \( 32 \times 10^5 \)
- Uncertainty in position (Δx) is similar for both the electron and helium.

2. **Understanding the Implication of Similar Uncertainty in Position:**
- Since the uncertainty in position (Δx) is the same for both the electron and helium, we can conclude that the uncertainty in momentum (Δp) will also be the same for both particles. This is because the uncertainty principle applies universally to all particles.

3. **Conclusion:**
- Therefore, the uncertainty in momentum of helium (Δp_He) will also be \( 32 \times 10^5 \).

### Final Answer:
The uncertainty in momentum of helium will be \( 32 \times 10^5 \).

---