An electron has a speed of 30,000 cm `"sec"^(-1)` accurate upto 0.001%. What is the uncertainty (in cm) in locating it's position?

Explanation

To solve the problem of determining the uncertainty in locating the position of an electron with a given speed, we will 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 (\(6.626 \times 10^{-34} \, \text{Js}\)).

### Step 1: Calculate the uncertainty in speed (\(\Delta v\))

Given that the speed of the electron is \(30,000 \, \text{cm/s}\) with an accuracy of \(0.001\%\), we first calculate the uncertainty in speed:

\[ \Delta v = \frac{0.001}{100} \times 30,000 \, \text{cm/s} = 0.3 \, \text{cm/s} \]

### Step 2: Calculate the uncertainty in momentum (\(\Delta p\))

The momentum \(p\) of the electron can be expressed as:

\[ p = mv \]

Thus, the uncertainty in momentum is given by:

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

The mass of the electron \(m\) is approximately \(9.1 \times 10^{-31} \, \text{kg}\). We first need to convert the speed uncertainty to SI units (meters per second):

\[ \Delta v = 0.3 \, \text{cm/s} = 0.3 \times 10^{-2} \, \text{m/s} \]

Now we can calculate \(\Delta p\):

\[ \Delta p = 9.1 \times 10^{-31} \, \text{kg} \cdot 0.3 \times 10^{-2} \, \text{m/s} = 2.73 \times 10^{-32} \, \text{kg m/s} \]

### Step 3: Use Heisenberg's uncertainty principle to find \(\Delta x\)

Now we can substitute \(\Delta p\) into the uncertainty principle equation:

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

Substituting the values:

\[ \Delta x \geq \frac{6.626 \times 10^{-34} \, \text{Js}}{4 \cdot 3.14 \cdot 2.73 \times 10^{-32} \, \text{kg m/s}} \]

Calculating the denominator:

\[ 4 \cdot 3.14 \cdot 2.73 \times 10^{-32} \approx 3.44 \times 10^{-31} \]

Now substituting back:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{3.44 \times 10^{-31}} \approx 1.93 \times 10^{-3} \, \text{m} \]

### Step 4: Convert \(\Delta x\) to centimeters

To convert meters to centimeters, we multiply by 100:

\[ \Delta x \approx 1.93 \times 10^{-3} \, \text{m} \times 100 = 0.193 \, \text{cm} \]

Thus, the uncertainty in locating the position of the electron is approximately:

\[ \Delta x \approx 0.193 \, \text{cm} \]

### Summary of the Solution
The uncertainty in locating the position of the electron is approximately **0.193 cm**.