A `alpha` -parhticle moves in a circular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. The de-Broglie wavelength assocaiated with the particle will be

Options

A. `1Ã…`
B. `0.1Ã…`
C. `10Ã…`
D. `0.01Ã…`

Explanation

To find the de-Broglie wavelength associated with an alpha particle moving in a circular path in a magnetic field, we can follow these steps:

### Step 1: Understand the relationship between forces
In a magnetic field, the centripetal force acting on the alpha particle is provided by the magnetic force. Therefore, we can equate the centripetal force to the magnetic force:

\[ F_{\text{centripetal}} = F_{\text{magnetic}} \]

This can be expressed as:

\[ \frac{m v^2}{r} = q v B \]

Where:
- \( m \) = mass of the alpha particle
- \( v \) = velocity of the alpha particle
- \( r \) = radius of the circular path
- \( q \) = charge of the alpha particle
- \( B \) = magnetic field strength

### Step 2: Solve for momentum
Rearranging the equation gives:

\[ m v = q v B r \]

Dividing both sides by \( v \) (assuming \( v \neq 0 \)):

\[ m = q B r \]

The momentum \( p \) of the particle is given by:

\[ p = m v = q B r v \]

### Step 3: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by:

\[ \lambda = \frac{h}{p} \]

Substituting the expression for momentum:

\[ \lambda = \frac{h}{q B r} \]

### Step 4: Substitute the known values
Now, we will substitute the known values into the equation. The constants are:

- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Charge of an alpha particle \( q = 2 \times 1.6 \times 10^{-19} \, \text{C} \)
- Magnetic field \( B = 0.25 \, \text{Wb/m}^2 \)
- Radius \( r = 0.83 \, \text{cm} = 0.0083 \, \text{m} \)

Substituting these values into the equation for wavelength:

\[ \lambda = \frac{6.63 \times 10^{-34}}{(2 \times 1.6 \times 10^{-19}) \times 0.25 \times 0.0083} \]

### Step 5: Calculate the wavelength
Calculating the denominator:

\[ q = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{C} \]

Now, substituting into the equation:

\[ \lambda = \frac{6.63 \times 10^{-34}}{(3.2 \times 10^{-19}) \times 0.25 \times 0.0083} \]

Calculating the denominator:

\[ 3.2 \times 10^{-19} \times 0.25 \times 0.0083 = 6.64 \times 10^{-22} \]

Now, substituting this back into the equation for \( \lambda \):

\[ \lambda = \frac{6.63 \times 10^{-34}}{6.64 \times 10^{-22}} \approx 9.96 \times 10^{-13} \, \text{m} \]

### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \( 10^{-10} \, \text{m} \)):

\[ \lambda \approx 9.96 \times 10^{-13} \, \text{m} = 9.96 \times 10^{1} \, \text{Å} \approx 0.01 \, \text{Å} \]

### Final Answer
Thus, the de-Broglie wavelength associated with the alpha particle is approximately:

\[ \lambda \approx 0.01 \, \text{Å} \]