Calculate the uncertainty in position of an electron whose velocity is `3.0 xx 10^4 cms^(-1)` accurate up to 0.001%. Mass of an electron `=9.1 xx 10^(-28)g`.

Explanation

To solve the problem of calculating the uncertainty in the position of an electron, we will use the Heisenberg uncertainty principle, which states:

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

where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta v\) is the uncertainty in velocity,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-28} \, \text{g}\)).

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

The velocity of the electron is given as \(3.0 \times 10^4 \, \text{cm/s}\) with an accuracy of \(0.001\%\).

To find the uncertainty in velocity, we calculate:

\[ \Delta v = \text{velocity} \times \frac{\text{accuracy}}{100} = 3.0 \times 10^4 \, \text{cm/s} \times \frac{0.001}{100} \]

Calculating this gives:

\[ \Delta v = 3.0 \times 10^4 \times 0.00001 = 0.30 \, \text{cm/s} \]

### Step 2: Substitute values into the Heisenberg uncertainty principle

Now, we will substitute the values into the uncertainty principle formula:

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

Substituting the known values:

- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 9.1 \times 10^{-28} \, \text{g} = 9.1 \times 10^{-31} \, \text{kg}\) (since \(1 \, \text{g} = 10^{-3} \, \text{kg}\))
- \(\Delta v = 0.30 \, \text{cm/s} = 0.30 \times 10^{-2} \, \text{m/s}\)

Now, we can calculate:

\[ \Delta x \cdot 0.30 \times 10^{-2} \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31}} \]

Calculating the denominator:

\[ 4 \times 3.14 \times 9.1 \times 10^{-31} \approx 1.136 \times 10^{-30} \]

Now, substituting this back into the equation:

\[ \Delta x \cdot 0.30 \times 10^{-2} \geq \frac{6.626 \times 10^{-34}}{1.136 \times 10^{-30}} \]

Calculating the right side:

\[ \frac{6.626 \times 10^{-34}}{1.136 \times 10^{-30}} \approx 5.83 \times 10^{-4} \]

### Step 3: Solve for \(\Delta x\)

Now, we can solve for \(\Delta x\):

\[ \Delta x \geq \frac{5.83 \times 10^{-4}}{0.30 \times 10^{-2}} = \frac{5.83 \times 10^{-4}}{3.0 \times 10^{-3}} \approx 0.194 \, \text{m} = 19.4 \, \text{cm} \]

### Conclusion

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

\[ \Delta x \geq 19.4 \, \text{cm} \]