Uncertainty in the position of an electron `("mass = "9.1 xx 10^(-31)kg)` moving with a velocity `300 ms^(-1)` accurate upto 0.001% will be

Options

A. `19.2 xx 10^(-2)m`
B. `5.76 xx 10^(-2)m`
C. `1.92 xx 10^(-2)m`
D. `3.84 xx 10^(-2)m`

Explanation

To find the uncertainty in the position of an electron moving with a given velocity, 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 (\(6.63 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)).

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

The velocity of the electron is given as \(300 \, \text{m/s}\) with an accuracy of \(0.001\%\).

To find the uncertainty in velocity, we calculate:

\[ \Delta v = \text{velocity} \times \left(\frac{\text{accuracy}}{100}\right) \]

Substituting the values:

\[ \Delta v = 300 \, \text{m/s} \times \left(\frac{0.001}{100}\right) = 300 \times 0.00001 = 0.003 \, \text{m/s} \]

### Step 2: Substitute values into the Heisenberg Uncertainty Principle formula

Now we can substitute the values into the Heisenberg Uncertainty Principle formula to find \(\Delta x\):

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

Rearranging gives:

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

### Step 3: Substitute known values

Substituting the known values:

- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(\Delta v = 0.003 \, \text{m/s}\)

Calculating:

\[ \Delta x \geq \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.003} \]

### Step 4: Calculate the denominator

Calculating the denominator:

\[ 4 \times 3.14 \approx 12.56 \]
\[ 12.56 \times 9.1 \times 10^{-31} \approx 1.144 \times 10^{-29} \]
\[ 1.144 \times 10^{-29} \times 0.003 \approx 3.432 \times 10^{-32} \]

### Step 5: Calculate \(\Delta x\)

Now substituting back into the equation:

\[ \Delta x \geq \frac{6.63 \times 10^{-34}}{3.432 \times 10^{-32}} \approx 1.93 \times 10^{-2} \, \text{m} \]

### Conclusion

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

\[ \Delta x \approx 1.93 \times 10^{-2} \, \text{m} \]