Assuming an electron is confined to a 1 nm wide region. Find the uncertainty in momentum using Heisenberg uncertainty principle.
`("Take h"=6.63xx10^(-34)Js)`

Options

A. `1.05xx10^(-25)"kg m s"^(-1)`
B. `2.03xx10^(-31)"kg m s"^(-1)`
C. `3.05xx10^(-34)"kg m s"^(-1)`
D. `3.05xx10^(-32)"kg m s"^(-1)`

Explanation

To find the uncertainty in momentum (\( \Delta P \)) for an electron confined to a 1 nm wide region using the Heisenberg uncertainty principle, we can follow these steps:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg uncertainty principle states that the product of the uncertainties in position (\( \Delta X \)) and momentum (\( \Delta P \)) is given by:
\[ \Delta X \cdot \Delta P \geq \frac{h}{4\pi} \]
In this case, we can use the simplified form:
\[ \Delta P \leq \frac{h}{\Delta X} \]
where \( h \) is Planck's constant.

### Step 2: Identify the values
Given:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- \( \Delta X = 1 \, \text{nm} = 1 \times 10^{-9} \, \text{m} \)

### Step 3: Substitute the values into the equation
Substituting the values into the equation for \( \Delta P \):
\[ \Delta P \leq \frac{6.63 \times 10^{-34}}{1 \times 10^{-9}} \]

### Step 4: Calculate \( \Delta P \)
Calculating the right side:
\[ \Delta P \leq \frac{6.63 \times 10^{-34}}{1 \times 10^{-9}} = 6.63 \times 10^{-25} \, \text{kg m/s} \]

### Step 5: Final calculation using \( 2\pi \)
To get the uncertainty in momentum more accurately, we can use:
\[ \Delta P \leq \frac{h}{2\pi \Delta X} \]
Substituting the values:
\[ \Delta P \leq \frac{6.63 \times 10^{-34}}{2\pi \times 1 \times 10^{-9}} \]

Calculating \( 2\pi \):
\[ 2\pi \approx 6.2832 \]
Now substituting this value:
\[ \Delta P \leq \frac{6.63 \times 10^{-34}}{6.2832 \times 10^{-9}} \approx 1.05 \times 10^{-25} \, \text{kg m/s} \]

### Final Result
Thus, the uncertainty in momentum (\( \Delta P \)) is approximately:
\[ \Delta P \approx 1.05 \times 10^{-25} \, \text{kg m/s} \]

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