Alpha perticles `(m = 6.7 xx 10^(-27)kg, q = +2e)` are accelerated from rest through a potential difference of 6.7 kV. Then they enter a magnetic field B = 0.2 T perpendicular to them direction of their motion. The radius of the path described by them is

Options

A. 8.375 m
B. 8.375 cm
C. infinity
D. none of these

Answer

To solve the problem, we need to find the radius of the path described by alpha particles after they are accelerated through a potential difference and enter a magnetic field. Here are the steps to arrive at the solution:

### Step 1: Calculate the Kinetic Energy
The kinetic energy (KE) gained by the alpha particles when accelerated through a potential difference \( V \) is given by the formula:
\[ KE = qV \]
where:
- \( q \) is the charge of the alpha particle,
- \( V \) is the potential difference.

Given:
- Charge of an alpha particle \( q = +2e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \),
- Potential difference \( V = 6.7 \, \text{kV} = 6.7 \times 10^3 \, \text{V} \).

Now, substituting the values:
\[ KE = (3.2 \times 10^{-19} \, \text{C}) \times (6.7 \times 10^3 \, \text{V}) = 2.144 \times 10^{-15} \, \text{J} \]

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of mass and velocity:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv^2 = qV \]
From this, we can solve for \( v \):
\[ v^2 = \frac{2qV}{m} \]
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 3: Substitute Values to Find Velocity
Substituting the known values:
- Mass of alpha particle \( m = 6.7 \times 10^{-27} \, \text{kg} \),
- Charge \( q = 3.2 \times 10^{-19} \, \text{C} \),
- Potential difference \( V = 6.7 \times 10^3 \, \text{V} \).

Calculating \( v \):
\[ v = \sqrt{\frac{2 \times (3.2 \times 10^{-19}) \times (6.7 \times 10^3)}{6.7 \times 10^{-27}}} \]
Calculating the numerator:
\[ = 2 \times 3.2 \times 10^{-19} \times 6.7 \times 10^3 = 4.2944 \times 10^{-15} \]
Now, dividing by the mass:
\[ \frac{4.2944 \times 10^{-15}}{6.7 \times 10^{-27}} = 6.404 \times 10^{11} \]
Taking the square root:
\[ v = \sqrt{6.404 \times 10^{11}} \approx 8.01 \times 10^5 \, \text{m/s} \]

### Step 4: Calculate the Radius of the Path
The radius \( r \) of the circular path in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]
Substituting the values:
- \( B = 0.2 \, \text{T} \),
- \( m = 6.7 \times 10^{-27} \, \text{kg} \),
- \( q = 3.2 \times 10^{-19} \, \text{C} \),
- \( v \approx 8.01 \times 10^5 \, \text{m/s} \).

Calculating \( r \):
\[ r = \frac{(6.7 \times 10^{-27}) \times (8.01 \times 10^5)}{(3.2 \times 10^{-19}) \times (0.2)} \]
Calculating the numerator:
\[ = 5.36967 \times 10^{-21} \]
Calculating the denominator:
\[ = 6.4 \times 10^{-20} \]
Thus:
\[ r \approx \frac{5.36967 \times 10^{-21}}{6.4 \times 10^{-20}} \approx 0.0839 \, \text{m} \approx 8.39 \, \text{cm} \]

### Final Answer
The radius of the path described by the alpha particles is approximately \( 8.39 \, \text{cm} \).

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