In an atom, an electron is moving with a speed of ` 600 m//s` with an accuracy of ` 0.005%` . Certainty with which the position of the electron can be localized is :
` ( h = 6.6 xx 10^(-34) kg m^2 s^(-1)` ,
mass of electron ` (e_m) = 9. 1 xx10^(-31) kg)`.

Options

A. `1.92xx10^(-3)m`
B. `3.84xx10^(-3)m`
C. `1.52xx10^(-4)m`
D. `5.10xx10^(-3)m`

Answer

To solve the problem of determining the certainty with which the position of an electron can be localized, we will use the Heisenberg Uncertainty Principle. Here are the steps to arrive at the solution:

### Step 1: Understand the Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle states that the product of the uncertainties in position (Δx) and momentum (Δp) of a particle is given by the equation:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
where \( h \) is Planck's constant.

### Step 2: Calculate the uncertainty in velocity (Δv)
The speed of the electron is given as \( 600 \, m/s \) with an accuracy of \( 0.005\% \). First, we need to calculate the uncertainty in velocity (Δv):
\[ \Delta v = \frac{0.005}{100} \times 600 = 0.03 \, m/s \]

### Step 3: Calculate the uncertainty in momentum (Δp)
The momentum (p) of the electron is given by:
\[ p = m \cdot v \]
where \( m \) is the mass of the electron. The uncertainty in momentum (Δp) can be calculated as:
\[ \Delta p = m \cdot \Delta v \]
Given:
- Mass of electron \( m = 9.1 \times 10^{-31} \, kg \)
- Δv from step 2 \( \Delta v = 0.03 \, m/s \)

Now, substituting the values:
\[ \Delta p = 9.1 \times 10^{-31} \, kg \cdot 0.03 \, m/s = 2.73 \times 10^{-32} \, kg \cdot m/s \]

### Step 4: Substitute Δp into the Heisenberg Uncertainty Principle
Now, we can substitute Δp into the Heisenberg Uncertainty Principle equation:
\[ \Delta x \cdot \Delta p = \frac{h}{4\pi} \]
Rearranging gives:
\[ \Delta x = \frac{h}{4\pi \Delta p} \]

### Step 5: Substitute values to find Δx
Using \( h = 6.626 \times 10^{-34} \, kg \cdot m^2/s \):
\[ \Delta x = \frac{6.626 \times 10^{-34}}{4 \cdot \pi \cdot 2.73 \times 10^{-32}} \]

Calculating the denominator:
\[ 4 \cdot \pi \cdot 2.73 \times 10^{-32} \approx 3.42 \times 10^{-31} \]

Now substituting back:
\[ \Delta x = \frac{6.626 \times 10^{-34}}{3.42 \times 10^{-31}} \approx 1.94 \times 10^{-3} \, m \]

### Final Answer
The certainty with which the position of the electron can be localized is approximately:
\[ \Delta x \approx 1.94 \times 10^{-3} \, m \]