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 the Heisenberg Uncertainty Principle, which states that the uncertainty in position (Δx) and the uncertainty in momentum (Δp) of a particle 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 (\(6.626 \times 10^{-34} \, \text{Js}\))

Given:
- The uncertainty in momentum of the electron (\( \Delta p_e \)) is \( 32 \times 10^5 \).

Since the problem states that the uncertainty in position of the electron and helium is similar, we can conclude that the uncertainty in momentum for helium (\( \Delta p_{He} \)) will also be the same as that of the electron.

Thus, we can write:

\[ \Delta p_{He} = \Delta p_e \]

Substituting the given value:

\[ \Delta p_{He} = 32 \times 10^5 \]

Therefore, the uncertainty in momentum of helium is:

\[ \Delta p_{He} = 32 \times 10^5 \]

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

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