What is the de - Broglie wavelength of the alpha - particle accelerated through a potential difference `V`?

Options

A. `(0.287)/(sqrt(V)) Å`
B. `(12.27)/(sqrt(V)) Å`
C. `(0.101)/(sqrtV)Å`
D. `(0.202)/(sqrt(V))Å`

Explanation

To find the de Broglie wavelength of an alpha particle accelerated through a potential difference \( V \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding de Broglie Wavelength**:
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 particle.

2. **Relating Momentum to Kinetic Energy**:
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( KE \):
\[ p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the particle.

3. **Kinetic Energy of the Alpha Particle**:
When an alpha particle (which has a charge \( Q \)) is accelerated through a potential difference \( V \), its kinetic energy is given by:
\[ KE = Q \cdot V \]
For an alpha particle, which consists of 2 protons, the charge \( Q \) is:
\[ Q = 2 \cdot e \quad \text{(where \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \, C \))} \]

4. **Substituting Kinetic Energy into Momentum**:
Now substituting \( KE \) into the momentum equation:
\[ p = \sqrt{2m \cdot (Q \cdot V)} = \sqrt{2m \cdot (2e \cdot V)} = \sqrt{4meV} \]

5. **Substituting Momentum into de Broglie Wavelength**:
Now substituting this expression for momentum back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{4meV}} = \frac{h}{2\sqrt{meV}} \]

6. **Calculating Mass of the Alpha Particle**:
The mass of an alpha particle (which is roughly the mass of 4 nucleons) is approximately:
\[ m \approx 4 \cdot (1.67 \times 10^{-27} \, kg) = 6.68 \times 10^{-27} \, kg \]

7. **Substituting Known Values**:
Now substituting \( h \) (Planck's constant, \( 6.626 \times 10^{-34} \, Js \)), \( m \), and \( e \) into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2 \sqrt{(6.68 \times 10^{-27})(1.6 \times 10^{-19})V}} \]

8. **Simplifying the Expression**:
Calculate the denominator:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2 \sqrt{(1.0688 \times 10^{-45})V}} = \frac{6.626 \times 10^{-34}}{2.065 \times 10^{-23} \sqrt{V}} \]
This simplifies to:
\[ \lambda = \frac{3.21 \times 10^{-11}}{\sqrt{V}} \, m \]

9. **Converting to Angstroms**:
Since \( 1 \, \text{m} = 10^{10} \, \text{Å} \):
\[ \lambda = \frac{3.21 \times 10^{-11} \times 10^{10}}{\sqrt{V}} = \frac{0.321}{\sqrt{V}} \, \text{Å} \]

### Final Answer
Thus, the de Broglie wavelength of the alpha particle accelerated through a potential difference \( V \) is:
\[ \lambda = \frac{0.321}{\sqrt{V}} \, \text{Å} \]