Calculate the wavelength associated with an electron moving with a velocity of `10^(3) m s^(-1)`.

Explanation

To calculate the wavelength associated with an electron moving with a velocity of \(10^3 \, \text{m/s}\), we will use the De Broglie wavelength formula:

\[ \lambda = \frac{h}{p} \]

where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant,
- \(p\) is the momentum of the electron.

### Step 1: Identify the values needed
1. **Planck's constant (h)**: \(6.63 \times 10^{-34} \, \text{Js}\)
2. **Mass of the electron (m)**: \(9.1 \times 10^{-31} \, \text{kg}\)
3. **Velocity of the electron (v)**: \(10^3 \, \text{m/s}\)

### Step 2: Calculate the momentum (p)
The momentum \(p\) of the electron can be calculated using the formula:

\[ p = m \cdot v \]

Substituting the values:

\[ p = (9.1 \times 10^{-31} \, \text{kg}) \cdot (10^3 \, \text{m/s}) = 9.1 \times 10^{-28} \, \text{kg m/s} \]

### Step 3: Substitute values into the De Broglie wavelength formula
Now we substitute \(h\) and \(p\) into the De Broglie wavelength formula:

\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{Js}}{9.1 \times 10^{-28} \, \text{kg m/s}} \]

### Step 4: Perform the division
Calculating the above expression:

\[ \lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-28}} = 6.63 \times 10^{-34} \times \frac{1}{9.1 \times 10^{-28}} = 6.63 \times 10^{-34} \times 1.0989 \times 10^{28} \]

This simplifies to:

\[ \lambda = 6.63 \times 10^{-6} \, \text{m} \]

### Step 5: Final calculation
Now, we can calculate the numerical value:

\[ \lambda \approx 0.72 \times 10^{-6} \, \text{m} = 0.72 \, \mu m \]

### Final Answer
The wavelength associated with the electron is approximately:

\[ \lambda \approx 0.72 \, \mu m \]