The uncertainty in the momentum of an electron is `1.0 xx 10^(-5) kgms^(-1)` . The uncertainty in its position will be : ( h = `6.626 xx 10^(-34)` Js )

Options

A. `1.05xx10^(-28) m`
B. `1.05xx10^(-26) m`
C. `5.27xx10^(-30) m`
D. `5.25xx10^(-28) m`

Answer

To find the uncertainty in the position of an electron given the uncertainty in its momentum, we can use Heisenberg's Uncertainty Principle. The formula we will use is:

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

Where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(h\) is Planck's constant.

Given:
- \(\Delta p = 1.0 \times 10^{-5} \, \text{kg m/s}\)
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)

### Step 1: Rearranging the formula
We need to isolate \(\Delta x\):
\[ \Delta x \geq \frac{h}{4\pi \Delta p} \]

### Step 2: Substitute the known values
Now, substitute the values of \(h\) and \(\Delta p\) into the equation:
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{4 \times \pi \times 1.0 \times 10^{-5}} \]

### Step 3: Calculate the denominator
First, calculate \(4 \times \pi\):
\[ 4 \times \pi \approx 4 \times 3.14 = 12.56 \]

Now, calculate \(4 \times \pi \times 1.0 \times 10^{-5}\):
\[ 4 \times \pi \times 1.0 \times 10^{-5} \approx 12.56 \times 10^{-5} = 1.256 \times 10^{-4} \]

### Step 4: Calculate \(\Delta x\)
Now, substitute this back into the equation for \(\Delta x\):
\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-4}} \]

### Step 5: Perform the division
Now, perform the division:
\[ \Delta x \geq 5.27 \times 10^{-30} \, \text{meters} \]

### Conclusion
Thus, the uncertainty in the position of the electron is approximately:
\[ \Delta x \approx 5.27 \times 10^{-30} \, \text{meters} \]

### Final Answer
The final answer is \(5.27 \times 10^{-30} \, \text{meters}\).

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