What is de-Broglie wavelength of the electron accelerated through a potential difference of 100V?

Options

A. `12.27Å`
B. `1.227Å`
C. `0.1227Å`
D. `0.001227Å`

Answer

To find the de-Broglie wavelength of an electron accelerated through a potential difference of 100V, we can use the de-Broglie wavelength formula:

\[ \lambda = \frac{h}{\sqrt{2 m e V}} \]

Where:
- \( \lambda \) is the de-Broglie wavelength,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( m \) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \( e \) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \( V \) is the potential difference (100V in this case).

### Step-by-Step Solution:

1. **Identify the constants**:
- Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Potential difference, \( V = 100 \, \text{V} \)

2. **Substitute the values into the formula**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100}} \]

3. **Calculate the denominator**:
- First calculate \( 2 \times 9.1 \times 10^{-31} \):
\[ 2 \times 9.1 \times 10^{-31} = 1.82 \times 10^{-30} \]
- Then calculate \( 1.6 \times 10^{-19} \times 100 \):
\[ 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \]
- Now multiply these results:
\[ 1.82 \times 10^{-30} \times 1.6 \times 10^{-17} = 2.912 \times 10^{-47} \]
- Finally, take the square root:
\[ \sqrt{2.912 \times 10^{-47}} \approx 5.39 \times 10^{-24} \]

4. **Calculate the de-Broglie wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.227 \times 10^{-10} \, \text{m} \]

5. **Convert to Angstroms**:
- Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 1.227 \, \text{Å} \]

### Final Answer:
The de-Broglie wavelength of the electron accelerated through a potential difference of 100V is approximately \( 1.227 \, \text{Å} \).