An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron becomes

Options

A. `1 Å`
B. `sqrt(1.5Å)`
C. `sqrt(3)Å`
D. `12.27Å`

Explanation

To find the de Broglie wavelength of an electron that has been accelerated through a potential difference, we can follow these steps:

### Step 1: Calculate the Additional Kinetic Energy Gained
The initial kinetic energy (KE_initial) of the electron is given as 100 eV. When the electron is accelerated through a potential difference of 50 V, it gains additional kinetic energy (KE_additional) equal to the charge of the electron multiplied by the potential difference.

\[ KE_{\text{additional}} = e \cdot V = 1.6 \times 10^{-19} \, \text{C} \times 50 \, \text{V} = 8.0 \times 10^{-18} \, \text{J} \]

### Step 2: Convert Initial Kinetic Energy to Joules
The initial kinetic energy in joules can be calculated as:

\[ KE_{\text{initial}} = 100 \, \text{eV} = 100 \times 1.6 \times 10^{-19} \, \text{J} = 1.6 \times 10^{-17} \, \text{J} \]

### Step 3: Calculate Total Kinetic Energy
Now, we can find the total kinetic energy (KE_total) of the electron after being accelerated:

\[ KE_{\text{total}} = KE_{\text{initial}} + KE_{\text{additional}} = 1.6 \times 10^{-17} \, \text{J} + 8.0 \times 10^{-18} \, \text{J} = 2.4 \times 10^{-17} \, \text{J} \]

### Step 4: Use the de Broglie Wavelength Formula
The de Broglie wavelength (λ) can be calculated using the formula:

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

Where:
- \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, \text{J s}) \)
- \( m \) is the mass of the electron \( (9.1 \times 10^{-31} \, \text{kg}) \)
- \( KE \) is the total kinetic energy calculated above.

Substituting the values:

\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.4 \times 10^{-17}}} \]

### Step 5: Calculate the Denominator
First, calculate the denominator:

\[ \sqrt{2 \times 9.1 \times 10^{-31} \times 2.4 \times 10^{-17}} = \sqrt{4.368 \times 10^{-47}} \approx 6.6 \times 10^{-24} \]

### Step 6: Calculate the Wavelength
Now substituting back into the wavelength formula:

\[ \lambda = \frac{6.63 \times 10^{-34}}{6.6 \times 10^{-24}} \approx 1.0 \times 10^{-10} \, \text{m} \]

### Final Answer
The de Broglie wavelength of the electron is approximately \( 1 \, \text{Å} \) (angstrom).