An `alpha-"particle"` is accelerated through a potential difference of `V` volts from rest. The de-Broglie's wavelengths associated with it is.

Options

A. `sqrt((150)/(V))A^(@)`
B. `(0.286)/sqrt(V)A^(@)`
C. `(0.101)/sqrt(V)A^(@)`
D. `(0.983)/sqrt(V)A^(@)`

Answer

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

### Step-by-Step Solution

1. **Understand the Energy Relationship**:
When an alpha particle is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy (KE) gained by the alpha particle can be expressed as:
\[ KE = qV \]
where \( q \) is the charge of the alpha particle. An alpha particle consists of 2 protons and 2 neutrons, so its charge \( q \) is \( 2e \) (where \( e \) is the elementary charge, approximately \( 1.602 \times 10^{-19} \) coulombs). Therefore:
\[ KE = 2eV \]

2. **Relate Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ 2eV = \frac{1}{2} mv^2 \]

3. **Solve for Velocity**:
Rearranging the equation to solve for \( v \):
\[ mv^2 = 4eV \implies v^2 = \frac{4eV}{m} \implies v = \sqrt{\frac{4eV}{m}} \]

4. **Use the de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting \( v \) from the previous step:
\[ \lambda = \frac{h}{m \sqrt{\frac{4eV}{m}}} \]
Simplifying this expression:
\[ \lambda = \frac{h}{\sqrt{4emV}} = \frac{h}{2\sqrt{emV}} \]

5. **Substitute Constants**:
Now substituting the values:
- Planck's constant \( h = 6.626 \times 10^{-34} \) J·s
- Mass of the alpha particle \( m \approx 4 \times 1.67 \times 10^{-27} \) kg (since an alpha particle is essentially a helium nucleus).
- Charge \( e \approx 1.602 \times 10^{-19} \) C.

Thus, the expression becomes:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2\sqrt{(1.602 \times 10^{-19})(4 \times 1.67 \times 10^{-27})V}} \]

6. **Final Expression**:
After simplifying, we can express the wavelength in terms of \( V \):
\[ \lambda \approx \frac{0.101}{\sqrt{V}} \text{ (in angstroms)} \]

### Final Answer:
The de Broglie wavelength associated with the alpha particle is:
\[ \lambda \approx \frac{0.101}{\sqrt{V}} \text{ angstroms} \]