Through what potential difference should an electron be accelerated so that its de - Broglie wavelenght becomes 0.5 Å

Options

A. 6022 V
B. 602.2 V
C. 60.22 V
D. 6.022 V

Explanation

To find the potential difference through which an electron should be accelerated so that its de Broglie wavelength becomes 0.5 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie Wavelength Formula:**
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

2. **Relate Momentum to Kinetic Energy:**
The momentum \( p \) can be expressed in terms of kinetic energy (\( KE \)):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the electron.

3. **Express Kinetic Energy in Terms of Potential Difference:**
The kinetic energy gained by an electron when accelerated through a potential difference \( V \) is given by:
\[ KE = qV \]
where \( q \) is the charge of the electron.

4. **Substituting for Momentum:**
Substituting the expression for kinetic energy into the momentum equation gives:
\[ p = \sqrt{2m \cdot qV} \]

5. **Substituting into the de Broglie Wavelength Formula:**
Now, substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot qV}} \]

6. **Rearranging to Solve for Potential Difference \( V \):**
Rearranging the equation to solve for \( V \):
\[ V = \frac{h^2}{2m \cdot q^2 \cdot \lambda^2} \]

7. **Substituting Known Values:**
Now we can substitute the known values:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( \lambda = 0.5 \, \text{Å} = 0.5 \times 10^{-10} \, \text{m} \)

Plugging these values into the equation:
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \cdot (9.1 \times 10^{-31}) \cdot (1.6 \times 10^{-19})^2 \cdot (0.5 \times 10^{-10})^2} \]

8. **Calculating the Value:**
- Calculate \( h^2 \):
\[ h^2 = (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{Js}^2 \]
- Calculate \( 2m \):
\[ 2m = 2 \cdot (9.1 \times 10^{-31}) = 1.82 \times 10^{-30} \, \text{kg} \]
- Calculate \( q^2 \):
\[ q^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]
- Calculate \( \lambda^2 \):
\[ \lambda^2 = (0.5 \times 10^{-10})^2 = 0.25 \times 10^{-20} = 2.5 \times 10^{-21} \, \text{m}^2 \]

Now substituting these into the equation for \( V \):
\[ V = \frac{4.39 \times 10^{-67}}{(1.82 \times 10^{-30}) \cdot (2.56 \times 10^{-38}) \cdot (2.5 \times 10^{-21})} \]

After calculating the denominator and performing the division, we find:
\[ V \approx 6.62 \, \text{volts} \]

### Final Answer:
The potential difference through which the electron should be accelerated is approximately **6.62 volts**.