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: Identify the given values
- Radius of the circular path, \( r = 0.83 \, \text{cm} = 0.0083 \, \text{m} \)
- Magnetic field strength, \( B = 0.25 \, \text{Wb/m}^2 \)
- Charge of an alpha particle, \( q = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)
### Step 2: Use the formula for the de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[
\lambda = \frac{h}{p}
\]
where \( p \) is the momentum of the particle.
### Step 3: Relate momentum to the magnetic force
In a magnetic field, the centripetal force is provided by the magnetic force:
\[
qvB = \frac{mv^2}{r}
\]
From this, we can express momentum \( p \) as:
\[
p = mv = qv \cdot r
\]
Thus, we can rewrite the wavelength formula as:
\[
\lambda = \frac{h}{qv \cdot r}
\]
### Step 4: Express velocity in terms of magnetic field and radius
From the equation \( qvB = \frac{mv^2}{r} \), we can rearrange to find \( v \):
\[
v = \frac{qBr}{m}
\]
Substituting this back into the wavelength formula gives:
\[
\lambda = \frac{h}{q \cdot \frac{qBr}{m} \cdot r} = \frac{h \cdot m}{q^2 B r^2}
\]
### Step 5: Calculate the mass of the alpha particle
The mass of an alpha particle is approximately \( m \approx 4 \times 1.67 \times 10^{-27} \, \text{kg} = 6.68 \times 10^{-27} \, \text{kg} \).
### Step 6: Substitute the values into the equation
Now substituting the values into the formula:
\[
\lambda = \frac{(6.63 \times 10^{-34} \, \text{Js}) \cdot (6.68 \times 10^{-27} \, \text{kg})}{(3.2 \times 10^{-19} \, \text{C})^2 \cdot (0.25 \, \text{Wb/m}^2) \cdot (0.0083 \, \text{m})^2}
\]
### Step 7: Calculate the wavelength
Calculating the denominator:
\[
(3.2 \times 10^{-19})^2 \cdot 0.25 \cdot (0.0083)^2
\]
Calculating this gives:
\[
= 1.024 \times 10^{-37} \cdot 0.25 \cdot 6.889 \times 10^{-5} \approx 1.76 \times 10^{-42}
\]
Now substituting into the wavelength formula:
\[
\lambda \approx \frac{(6.63 \times 10^{-34}) \cdot (6.68 \times 10^{-27})}{1.76 \times 10^{-42}} \approx 2.51 \times 10^{-12} \, \text{m}
\]
### Step 8: Convert to Angstroms
To convert meters to Angstroms:
\[
\lambda \approx 2.51 \times 10^{-12} \, \text{m} = 25.1 \, \text{pm} = 0.251 \, \text{Å}
\]
### Final Answer
The de-Broglie wavelength associated with the alpha particle is approximately \( 0.251 \, \text{Å} \).
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