The uncertainty in the position of an electron `(mass = 9.1 xx 10^-28 g)` moving with a velocity of `3.0 xx 10^4 cm s^-1` accurate up to `0.001 %` will be
(Use `(h)/(4 pi)` in the uncertainty expression, where `h = 6.626 xx 10^-27 erg - s`)

Options

A. `1.92 cm`
B. `7.68 cm`
C. `5.76 cm`
D. `3.84 cm`

Explanation

To find the uncertainty in the position of an electron moving with a given velocity, we can use Heisenberg's uncertainty principle, which states:

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

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

### Step 1: Calculate \(\Delta v\)
Given that the velocity of the electron is \(3.0 \times 10^4 \, \text{cm/s}\) and the accuracy is \(0.001\%\), we can calculate the uncertainty in velocity (\(\Delta v\)).

\[ \Delta v = \text{Velocity} \times \frac{\text{Accuracy}}{100} = 3.0 \times 10^4 \, \text{cm/s} \times \frac{0.001}{100} = 3.0 \times 10^4 \, \text{cm/s} \times 0.00001 = 3.0 \times 10^{-1} \, \text{cm/s} \]

### Step 2: Calculate \(\Delta p\)
The uncertainty in momentum (\(\Delta p\)) is given by:

\[ \Delta p = m \cdot \Delta v \]

Where \(m\) is the mass of the electron, given as \(9.1 \times 10^{-28} \, \text{g}\).

Substituting the values:

\[ \Delta p = 9.1 \times 10^{-28} \, \text{g} \cdot 3.0 \times 10^{-1} \, \text{cm/s} = 2.73 \times 10^{-28} \, \text{g cm/s} \]

### Step 3: Substitute into the uncertainty principle
Now we can substitute \(\Delta p\) into the uncertainty principle equation to find \(\Delta x\):

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

Substituting \(h = 6.626 \times 10^{-27} \, \text{erg s}\) and \(\Delta p = 2.73 \times 10^{-28} \, \text{g cm/s}\):

\[ \Delta x \geq \frac{6.626 \times 10^{-27}}{4 \cdot \pi \cdot 2.73 \times 10^{-28}} \]

Calculating \(4\pi\):

\[ 4\pi \approx 12.566 \]

Now substituting this value:

\[ \Delta x \geq \frac{6.626 \times 10^{-27}}{12.566 \cdot 2.73 \times 10^{-28}} = \frac{6.626 \times 10^{-27}}{3.42 \times 10^{-27}} \approx 1.93 \, \text{cm} \]

### Final Result
Thus, the uncertainty in the position of the electron is approximately:

\[ \Delta x \approx 1.93 \, \text{cm} \]