An `alpha-`particle is accelerated by a potential difference of `10^4 V.` Find the change in its direction of motion, if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 tesla. (Given: mass of `alpha-`particle `6.4xx10^(-27)`kg).

Explanation

To solve the problem of an alpha particle accelerated by a potential difference and then entering a magnetic field, we can follow these steps:

### Step 1: Determine the charge and mass of the alpha particle
An alpha particle consists of 2 protons and 2 neutrons. Thus, its charge \( q \) is:
\[ q = 2 \times e = 2 \times 1.6 \times 10^{-19} \, \text{C} = 3.2 \times 10^{-19} \, \text{C} \]
The mass \( m \) of the alpha particle is given as:
\[ m = 6.4 \times 10^{-27} \, \text{kg} \]

### Step 2: Calculate the kinetic energy gained by the alpha particle
The kinetic energy \( KE \) gained by the alpha particle when accelerated through a potential difference \( V \) is given by:
\[ KE = qV \]
Substituting the values:
\[ KE = (3.2 \times 10^{-19} \, \text{C})(10^4 \, \text{V}) = 3.2 \times 10^{-15} \, \text{J} \]

### Step 3: Relate kinetic energy to velocity
The kinetic energy can also be expressed in terms of the velocity \( v \):
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ 3.2 \times 10^{-15} = \frac{1}{2} (6.4 \times 10^{-27}) v^2 \]
Solving for \( v^2 \):
\[ v^2 = \frac{2 \times 3.2 \times 10^{-15}}{6.4 \times 10^{-27}} = \frac{6.4 \times 10^{-15}}{6.4 \times 10^{-27}} = 10^{12} \]
Thus,
\[ v = \sqrt{10^{12}} = 10^6 \, \text{m/s} \]

### Step 4: Calculate the radius of the circular path in the magnetic field
The radius \( r \) of the circular path of the particle in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]
Substituting the known values:
\[ r = \frac{(6.4 \times 10^{-27})(10^6)}{(3.2 \times 10^{-19})(0.1)} = \frac{6.4 \times 10^{-21}}{3.2 \times 10^{-20}} = 0.2 \, \text{m} \]

### Step 5: Calculate the angle of deflection
The thickness of the magnetic field region is \( L = 0.1 \, \text{m} \). The angle \( \theta \) can be found using the relationship:
\[ \sin \theta = \frac{L}{r} \]
Substituting the known values:
\[ \sin \theta = \frac{0.1}{0.2} = 0.5 \]
Thus,
\[ \theta = \sin^{-1}(0.5) = 30^\circ \]

### Final Answer
The change in the direction of motion of the alpha particle is \( 30^\circ \).

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