The de-Broglie wavelength of an `alpha-`particles at a voltage V is
(Given that `alpha-` particle has 2 units positive charge and 4 units mass )

Options

A. `lambda = ( 12.3)/(sqrt(V))Å`
B. `lambda = ( 0.286)/(sqrt(V))Å`
C. `lambda = ( 0.101)/( sqrt( V))Å`
D. `lambda = ( 0.856)/(sqrt( V))Å`

Answer

To find the de-Broglie wavelength of an alpha particle at a voltage \( V \), we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and voltage
The kinetic energy (\( KE \)) gained by a charged particle when it is accelerated through a potential difference \( V \) is given by:
\[ KE = Q \cdot V \]
where \( Q \) is the charge of the particle.

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum (\( p \)):
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle.

### Step 3: Equate the two expressions for kinetic energy
Setting the two expressions for kinetic energy equal gives:
\[ Q \cdot V = \frac{p^2}{2m} \]

### Step 4: Solve for momentum
Rearranging the equation to solve for momentum \( p \):
\[ p^2 = 2m \cdot Q \cdot V \]
Taking the square root:
\[ p = \sqrt{2m \cdot Q \cdot V} \]

### Step 5: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting the expression for momentum:
\[ \lambda = \frac{h}{\sqrt{2m \cdot Q \cdot V}} \]

### Step 6: Substitute the values for the alpha particle
Given:
- The charge of the alpha particle \( Q = 2 \times 1.6 \times 10^{-19} \, \text{C} \)
- The mass of the alpha particle \( m = 4 \times 1.66 \times 10^{-27} \, \text{kg} \) (where 1 atomic mass unit is approximately \( 1.66 \times 10^{-27} \, \text{kg} \))
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)

### Step 7: Substitute the values into the wavelength formula
Substituting these values into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot (4 \times 1.66 \times 10^{-27}) \cdot (2 \cdot 1.6 \times 10^{-19}) \cdot V}} \]

### Step 8: Simplify the expression
Calculating the constants:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 4 \cdot 1.66 \times 10^{-27} \cdot 2 \cdot 1.6 \times 10^{-19} \cdot V}} \]
This will yield:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 4 \cdot 1.66 \cdot 2 \cdot 1.6 \cdot 10^{-46} \cdot V}} \]

### Step 9: Final expression for wavelength
After calculating the constants, we arrive at:
\[ \lambda = \frac{1.01 \times 10^{-10}}{\sqrt{V}} \, \text{meters} \]
Converting to angstroms (1 meter = \( 10^{10} \) angstroms):
\[ \lambda = 0.101 \, \text{Å} \cdot \frac{1}{\sqrt{V}} \]

### Conclusion
Thus, the de-Broglie wavelength of an alpha particle at a voltage \( V \) is:
\[ \lambda = \frac{0.101}{\sqrt{V}} \, \text{Å} \]