For an alpha particle , accelerated through a potential difference V , wavelength (in ` Å` ) of the associated matter wave is

Options

A. `(12.27)/(sqrt(V))`
B. `(0.101)/(sqrt(V))`
C. `(0.202)/(sqrt(V))`
D. `(0.286)/(sqrt(V))`

Answer

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

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
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 \( K.E. \) can be expressed as:
\[ K.E. = 2eV \]
where \( e \) is the charge of the alpha particle.

2. **Momentum Relation**:
The kinetic energy can also be expressed in terms of momentum \( p \) and mass \( m \):
\[ K.E. = \frac{p^2}{2m} \]
Equating the two expressions for kinetic energy gives:
\[ \frac{p^2}{2m} = 2eV \]

3. **Solving for Momentum**:
Rearranging the equation to solve for \( p^2 \):
\[ p^2 = 4m \cdot eV \]
Since the mass of an alpha particle is approximately \( 4 \times m_p \) (where \( m_p \) is the mass of a proton), we have:
\[ p^2 = 4(4m_p)eV = 16m_p \cdot eV \]

4. **Finding Momentum**:
Taking the square root gives:
\[ p = 4\sqrt{m_p \cdot eV} \]

5. **Using de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
Substituting the expression for \( p \):
\[ \lambda = \frac{h}{4\sqrt{m_p \cdot eV}} \]

6. **Substituting Values**:
The Planck's constant \( h \) is \( 6.63 \times 10^{-34} \, \text{J s} \) and the mass of a proton \( m_p \) is approximately \( 1.67 \times 10^{-27} \, \text{kg} \). Thus,
\[ \lambda = \frac{6.63 \times 10^{-34}}{4\sqrt{1.67 \times 10^{-27} \cdot V}} \]

7. **Converting Wavelength to Angstroms**:
Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \), we convert the wavelength to angstroms:
\[ \lambda \text{ (in Å)} = \frac{6.63 \times 10^{-34}}{4\sqrt{1.67 \times 10^{-27} \cdot V}} \times 10^{10} \]

8. **Final Expression**:
After simplification, we can express the wavelength in terms of \( V \):
\[ \lambda = \frac{0.101}{\sqrt{V}} \text{ Å} \]

### Final Answer:
The wavelength of the associated matter wave for an alpha particle accelerated through a potential difference \( V \) is:
\[ \lambda = \frac{0.101}{\sqrt{V}} \text{ Å} \]