Calculate de Broglie wavelength of an electron travelling ar `1 %` of the speed of light.

Options

A. `2.73 xx 10^-24`
B. `2.42 xx 10^-10`
C. `242.2 xx 10^10`
D. None of these

Explanation

To calculate the de Broglie wavelength of an electron traveling at 1% of the speed of light, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the speed of the electron**:
- The speed of light (c) is approximately \(3 \times 10^8\) m/s.
- Since the electron is traveling at 1% of the speed of light:
\[ v = \frac{1}{100} \times c = \frac{1}{100} \times 3 \times 10^8 \text{ m/s} = 3 \times 10^6 \text{ m/s} \]

2. **Use the de Broglie wavelength formula**:
- The de Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
- Where:
- \(h\) is Planck's constant, approximately \(6.626 \times 10^{-34}\) J·s.
- \(m\) is the mass of the electron, approximately \(9.1 \times 10^{-31}\) kg.
- \(v\) is the velocity of the electron, which we calculated as \(3 \times 10^6\) m/s.

3. **Substitute the values into the formula**:
\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J·s}}{(9.1 \times 10^{-31} \text{ kg}) \times (3 \times 10^6 \text{ m/s})} \]

4. **Calculate the denominator**:
- First, calculate the product of mass and velocity:
\[ mv = 9.1 \times 10^{-31} \text{ kg} \times 3 \times 10^6 \text{ m/s} = 2.73 \times 10^{-24} \text{ kg·m/s} \]

5. **Calculate the wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.73 \times 10^{-24}} \approx 2.424 \times 10^{-10} \text{ m} \]

6. **Final Result**:
- The de Broglie wavelength of the electron traveling at 1% of the speed of light is approximately:
\[ \lambda \approx 2.424 \times 10^{-10} \text{ m} \]