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 (λ) of an alpha particle accelerated through a potential difference (V), we can use the de Broglie wavelength formula for matter waves. The formula is given by:

\[ \lambda = \frac{h}{\sqrt{2 m e V}} \]

where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( e \) is the charge of the particle,
- \( V \) is the potential difference.

### Step 1: Identify the mass and charge of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. Therefore:
- The mass of the alpha particle \( m \) is approximately \( 4 \times m_p \) (where \( m_p \) is the mass of a proton).
- The charge of the alpha particle \( e \) is \( 2 \times e_p \) (where \( e_p \) is the charge of a proton).

### Step 2: Substitute values for mass and charge
Using the known values:
- Mass of a proton \( m_p = 1.67 \times 10^{-27} \) kg,
- Charge of a proton \( e_p = 1.6 \times 10^{-19} \) C.

Thus:
\[ m = 4 \times 1.67 \times 10^{-27} \text{ kg} = 6.68 \times 10^{-27} \text{ kg} \]
\[ e = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C} \]

### Step 3: Substitute values into the wavelength formula
Now substituting the values into the wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2 m e V}} = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (6.68 \times 10^{-27}) \times (3.2 \times 10^{-19}) \times V}} \]

### Step 4: Simplify the expression
Calculating the denominator:
\[ \sqrt{2 \times 6.68 \times 10^{-27} \times 3.2 \times 10^{-19}} = \sqrt{4.2688 \times 10^{-45}} = 6.53 \times 10^{-23} \]
Thus:
\[ \lambda = \frac{6.63 \times 10^{-34}}{6.53 \times 10^{-23} \sqrt{V}} \]

### Step 5: Final expression for wavelength
This simplifies to:
\[ \lambda = \frac{0.101}{\sqrt{V}} \times 10^{-10} \text{ m} \]
Converting to angstroms (1 Å = \( 10^{-10} \) m):
\[ \lambda = \frac{0.101}{\sqrt{V}} \text{ Å} \]

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