An electron is accelerated through a potential difference of `10,000V`. Its de Broglie wavelength is, (nearly): `(me=9xx10^(-31)kg)`

Options

A. 12.2nm
B. `12.2xx10^(-13)m`
C. `12.2xx10^(-12)m`
D. `12.2xx10^(-14)m`

Explanation

To find the de Broglie wavelength of an electron accelerated through a potential difference of 10,000 V, we can use the formula for the de Broglie wavelength:

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

where \( h \) is Planck's constant and \( p \) is the momentum of the electron. However, we can also use a simplified formula that relates the de Broglie wavelength to the potential difference:

\[ \lambda = \frac{12.27}{\sqrt{V}} \text{ (in angstroms)} \]

Given that \( V = 10,000 \, \text{V} \), we can substitute this value into the formula.

### Step-by-Step Solution:

1. **Identify the formula**: We will use the formula for the de Broglie wavelength in terms of the potential difference:
\[ \lambda = \frac{12.27}{\sqrt{V}} \]

2. **Substitute the value of V**: Here, \( V = 10,000 \, \text{V} \):
\[ \lambda = \frac{12.27}{\sqrt{10,000}} \]

3. **Calculate the square root of V**:
\[ \sqrt{10,000} = 100 \]

4. **Substitute back into the equation**:
\[ \lambda = \frac{12.27}{100} \]

5. **Perform the division**:
\[ \lambda = 0.1227 \, \text{angstroms} \]

6. **Convert angstroms to meters**: Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{meters} \):
\[ \lambda = 0.1227 \times 10^{-10} \, \text{meters} = 1.227 \times 10^{-12} \, \text{meters} \]

### Final Answer:
The de Broglie wavelength of the electron is approximately:
\[ \lambda \approx 1.227 \times 10^{-12} \, \text{meters} \]