If the uncertainty in the position of an electron is `10^(-10)` m, then what be the value of uncertainty in its momentum in kg `m s^(-1)` ? `(h = 6.62 xx10^(-34) Js)`

Options

A. `0.52xx10^(-24)`
B. `1.01xx10^(-24)`
C. `1.09xx10^(-24)`
D. `1.07xx10^(-24)`

Answer

To find the uncertainty in momentum (Δp) of an electron given the uncertainty in its position (Δx), we can use Heisenberg's Uncertainty Principle. The principle states that:

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

Where:
- Δx is the uncertainty in position
- Δp is the uncertainty in momentum
- h is Planck's constant

Given:
- Δx = \(10^{-10}\) m
- h = \(6.62 \times 10^{-34}\) Js

We can rearrange the equation to solve for Δp:

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

Now, let's calculate Δp step by step.

### Step 1: Calculate the value of \(4\pi\)

Using the approximate value of \(\pi \approx 3.14\):

\[ 4\pi \approx 4 \times 3.14 = 12.56 \]

### Step 2: Substitute the values into the equation

Now we can substitute h and Δx into the equation:

\[ \Delta p = \frac{6.62 \times 10^{-34}}{12.56 \times 10^{-10}} \]

### Step 3: Perform the calculation

Calculating the denominator:

\[ 12.56 \times 10^{-10} = 1.256 \times 10^{-9} \]

Now we can perform the division:

\[ \Delta p = \frac{6.62 \times 10^{-34}}{1.256 \times 10^{-9}} \approx 5.27 \times 10^{-25} \text{ kg m/s} \]

### Final Result

Thus, the uncertainty in momentum (Δp) is approximately:

\[ \Delta p \approx 5.27 \times 10^{-25} \text{ kg m/s} \]