The de borglie wavelength of an electron traveling with 10% of the velocity of light is

Options

A. 24.2 pm
B. 242.2 pm
C. 2.42 pm
D. 2.424 pm

Answer

To find the de Broglie wavelength of an electron traveling at 10% of the velocity of light, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda_d = \frac{h}{mv} \]
where:
- \( h \) = Planck's constant
- \( m \) = mass of the electron
- \( v \) = velocity of the electron

2. **Identify the Values**:
- The velocity of light \( c \) is approximately \( 3 \times 10^8 \) m/s.
- The electron is traveling at 10% of the velocity of light:
\[ v = 0.1c = 0.1 \times 3 \times 10^8 \text{ m/s} = 3 \times 10^7 \text{ m/s} \]
- The mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \) kg.
- Planck's constant \( h \) is approximately \( 6.626 \times 10^{-34} \) J·s.

3. **Substitute the Values into the Formula**:
Now, substitute the values into the de Broglie wavelength formula:
\[ \lambda_d = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(3 \times 10^7)} \]

4. **Calculate the Denominator**:
First, calculate the denominator:
\[ m \cdot v = (9.1 \times 10^{-31} \text{ kg})(3 \times 10^7 \text{ m/s}) = 2.73 \times 10^{-23} \text{ kg m/s} \]

5. **Calculate the Wavelength**:
Now, substitute back to find \( \lambda_d \):
\[ \lambda_d = \frac{6.626 \times 10^{-34}}{2.73 \times 10^{-23}} \approx 2.43 \times 10^{-11} \text{ m} \]

6. **Convert to Picometers**:
Since 1 picometer (pm) = \( 10^{-12} \) m, convert the wavelength:
\[ \lambda_d = 2.43 \times 10^{-11} \text{ m} = 24.3 \text{ pm} \]

7. **Final Answer**:
The de Broglie wavelength of the electron traveling with 10% of the velocity of light is approximately:
\[ \lambda_d \approx 24.3 \text{ pm} \]