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. 2 eV
B. 1 eV
C. 4 eV
D. `(1)/(2)` eV

Answer

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

### Step 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 particle.

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

### Step 3: Express kinetic energy in terms of potential difference
When an alpha particle is accelerated through a potential difference \( V \), its kinetic energy is given by:
\[ KE = Q \cdot V \]
where \( Q \) is the charge of the alpha particle. For an alpha particle, which consists of 2 protons and 2 neutrons, the charge \( Q \) is:
\[ Q = 2e = 2 \times 1.6 \times 10^{-19} \, \text{C} \]

### Step 4: Substitute kinetic energy into the momentum formula
Substituting \( KE \) into the momentum equation, we have:
\[ p = \sqrt{2m \cdot (Q \cdot V)} = \sqrt{2m \cdot (2e \cdot V)} \]

### Step 5: Substitute momentum into the de Broglie wavelength formula
Now, substituting \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot (2e \cdot V)}} \]

### Step 6: Calculate the mass of the alpha particle
The mass of an alpha particle (which is essentially a helium nucleus) is approximately:
\[ m = 4 \, \text{amu} = 4 \times 1.66 \times 10^{-27} \, \text{kg} \approx 6.64 \times 10^{-27} \, \text{kg} \]

### Step 7: Substitute values into the equation
Now substituting the values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot (6.64 \times 10^{-27}) \cdot (2 \cdot 1.6 \times 10^{-19} \cdot V)}} \]

### Step 8: Simplify the expression
This simplifies to:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \cdot 6.64 \times 10^{-27} \cdot 3.2 \times 10^{-19} \cdot V}} \]

### Step 9: Final expression for de Broglie wavelength
After calculating the constants and simplifying, we find:
\[ \lambda \approx \frac{0.101}{\sqrt{V}} \, \text{angstroms} \]

### Conclusion
Thus, the de Broglie wavelength associated with the alpha particle accelerated through a potential difference \( V \) volts is:
\[ \lambda \approx \frac{0.101}{\sqrt{V}} \, \text{angstroms} \]