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 with 10% of the velocity of light, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)),
- \( v \) is the velocity of the electron.

### Step 2: Calculate the velocity of the electron
The velocity of light (c) is approximately \( 3 \times 10^8 \, \text{m/s} \). Since the electron is traveling at 10% of the velocity of light, we calculate:
\[ v = 0.1 \times c = 0.1 \times 3 \times 10^8 \, \text{m/s} = 3 \times 10^7 \, \text{m/s} \]

### Step 3: Substitute the values into the de Broglie wavelength formula
Now we can substitute the values of \( h \), \( m \), and \( v \) into the formula:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(9.1 \times 10^{-31} \, \text{kg})(3 \times 10^7 \, \text{m/s})} \]

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

### Step 5: Calculate the de Broglie wavelength
Now substitute the denominator back into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.73 \times 10^{-23}} \approx 2.43 \times 10^{-11} \, \text{m} \]

### Step 6: Convert to picometers
To convert meters to picometers (1 pm = \( 10^{-12} \, \text{m} \)):
\[ \lambda \approx 2.43 \times 10^{-11} \, \text{m} = 24.3 \, \text{pm} \]

### Final Answer
The de Broglie wavelength of the electron traveling with 10% of the velocity of light is approximately:
\[ \lambda \approx 24.3 \, \text{pm} \]

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