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. `0.01 Å`
B. `1 Å`
C. `0.1 Å`
D. `10 Å`

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 magnetic force and centripetal force
The magnetic force acting on the charged particle provides the necessary centripetal force to keep it in circular motion. The magnetic force \( F \) is given by:
\[ F = QVB \]
where:
- \( Q \) is the charge of the particle,
- \( V \) is the velocity of the particle,
- \( B \) is the magnetic field strength.

The centripetal force \( F_c \) required to keep the particle in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the particle,
- \( r \) is the radius of the circular path.

### Step 2: Set the magnetic force equal to the centripetal force
Since the magnetic force provides the centripetal force, we can set these two equations equal to each other:
\[ QVB = \frac{mv^2}{r} \]

### Step 3: Solve for velocity \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{QBr}{m} \]

### Step 4: Calculate the de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle, given by \( p = mv \). Thus,
\[ \lambda = \frac{h}{mv} \]
Substituting for \( V \) from Step 3:
\[ \lambda = \frac{h}{m \cdot \frac{QBr}{m}} = \frac{h}{QBr} \]

### Step 5: Substitute known values
Now we need to substitute the known values:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Charge of an alpha particle \( Q = 2 \times 1.602 \times 10^{-19} \, \text{C} \)
- 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:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(2 \times 1.602 \times 10^{-19}) \cdot (0.25) \cdot (0.0083)} \]

### Step 6: Calculate the wavelength
Calculating the denominator:
\[ QBr = (2 \times 1.602 \times 10^{-19}) \cdot (0.25) \cdot (0.0083) = 2.6525 \times 10^{-24} \]
Now, substituting back to find \( \lambda \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.6525 \times 10^{-24}} \approx 2.50 \times 10^{-10} \, \text{m} \]

### Final Answer
The de-Broglie wavelength associated with the alpha particle is approximately \( 2.50 \times 10^{-10} \, \text{m} \) or \( 0.250 \, \text{nm} \).

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