The de-Broglie waves associated with an electron accelerated through a potential difference of 121V is :

Options

A. 1.227 nm
B. 12.270 nm
C. 0.112 nm
D. 11.200 nm

Answer

To find the de-Broglie wavelength associated with an electron accelerated through a potential difference of 121 V, 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:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Calculate the Momentum (p)**:
The momentum of an electron accelerated through a potential difference \( V \) can be expressed as:
\[ p = \sqrt{2m_e \cdot eV} \]
where:
- \( m_e \) is the mass of the electron (approximately \( 9.11 \times 10^{-31} \) kg),
- \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) C),
- \( V \) is the potential difference (121 V in this case).

3. **Substituting Values**:
First, we need to calculate \( p \):
\[ p = \sqrt{2 \cdot (9.11 \times 10^{-31} \text{ kg}) \cdot (1.6 \times 10^{-19} \text{ C}) \cdot (121 \text{ V})} \]

4. **Calculate the Value**:
- Calculate \( 2 \cdot m_e \cdot e \cdot V \):
\[ 2 \cdot (9.11 \times 10^{-31}) \cdot (1.6 \times 10^{-19}) \cdot (121) \approx 3.52 \times 10^{-48} \text{ kg} \cdot \text{m}^2/\text{s}^2 \]
- Now, take the square root to find \( p \):
\[ p \approx \sqrt{3.52 \times 10^{-48}} \approx 5.93 \times 10^{-24} \text{ kg m/s} \]

5. **Calculate the de-Broglie Wavelength**:
Now substitute \( p \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h \approx 6.63 \times 10^{-34} \text{ J s} \):
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{5.93 \times 10^{-24}} \approx 1.12 \times 10^{-10} \text{ m} \]

6. **Convert to Angstroms**:
Since \( 1 \text{ Angstrom} = 10^{-10} \text{ m} \):
\[ \lambda \approx 1.12 \text{ Angstroms} \]

7. **Convert to Nanometers**:
To convert Angstroms to nanometers (1 Angstrom = 0.1 nm):
\[ \lambda \approx 0.112 \text{ nm} \]

### Final Answer:
The de-Broglie wavelength associated with an electron accelerated through a potential difference of 121 V is approximately **0.112 nm**.