An `alpha`-particle and a proton having same kineti energy enetrs in uniform magnetic field perpendicularly. Let x be the ratio of their magnitude of accelation any y the ratio of their time periods. Then

Options

A. `x=(1)/((2)^(3//2))`
B. `x=(1)/(4)`
C. `y=2`
D. `y=4`

Answer

To solve the problem, we need to find the ratios of the magnitudes of acceleration (x) and time periods (y) of an alpha particle and a proton when both enter a uniform magnetic field perpendicularly with the same kinetic energy.

### Step 1: Understanding Kinetic Energy
Both the alpha particle and the proton have the same kinetic energy (KE). The kinetic energy is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \(m\) is the mass and \(v\) is the velocity of the particle.

### Step 2: Expressing Kinetic Energy for Both Particles
Let:
- \(m_p\) and \(v_p\) be the mass and velocity of the proton.
- \(m_{\alpha}\) and \(v_{\alpha}\) be the mass and velocity of the alpha particle.

Since both have the same kinetic energy:
\[ \frac{1}{2} m_p v_p^2 = \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]
This implies:
\[ m_p v_p^2 = m_{\alpha} v_{\alpha}^2 \]

### Step 3: Finding the Radius of Circular Motion
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path. The radius \(r\) of this path is given by:
\[ r = \frac{mv}{qB} \]
where \(q\) is the charge of the particle and \(B\) is the magnetic field strength.

### Step 4: Finding the Acceleration
The centripetal acceleration \(a\) of a particle moving in a circle is given by:
\[ a = \frac{v^2}{r} \]
Substituting the expression for \(r\):
\[ a = \frac{v^2 qB}{mv} = \frac{vqB}{m} \]

### Step 5: Finding the Ratio of Accelerations (x)
Now, we can find the ratio of the magnitudes of acceleration for the alpha particle and the proton:
\[ x = \frac{a_{\alpha}}{a_p} = \frac{v_{\alpha} q_{\alpha} B / m_{\alpha}}{v_p q_p B / m_p} \]
Since both particles have the same kinetic energy, we can express their velocities in terms of their masses:
\[ v_{\alpha} = \sqrt{\frac{2 KE}{m_{\alpha}}}, \quad v_p = \sqrt{\frac{2 KE}{m_p}} \]
Substituting these into the acceleration ratio:
\[ x = \frac{\sqrt{\frac{2 KE}{m_{\alpha}}} \cdot q_{\alpha} \cdot B / m_{\alpha}}{\sqrt{\frac{2 KE}{m_p}} \cdot q_p \cdot B / m_p} \]
The \(B\) and \(KE\) terms cancel out, leading to:
\[ x = \frac{q_{\alpha} m_p}{q_p m_{\alpha}} \cdot \sqrt{\frac{m_p}{m_{\alpha}}} \]

### Step 6: Finding the Time Period (y)
The time period \(T\) for circular motion is given by:
\[ T = \frac{2\pi r}{v} \]
Substituting for \(r\):
\[ T = \frac{2\pi mv}{qB} \]
Thus, the ratio of the time periods is:
\[ y = \frac{T_{\alpha}}{T_p} = \frac{m_{\alpha} v_{\alpha}/(q_{\alpha} B)}{m_p v_p/(q_p B)} = \frac{m_{\alpha} v_{\alpha} q_p}{m_p v_p q_{\alpha}} \]
Using the expressions for velocities:
\[ y = \frac{m_{\alpha} \sqrt{\frac{2 KE}{m_{\alpha}}} \cdot q_p}{m_p \sqrt{\frac{2 KE}{m_p}} \cdot q_{\alpha}} \]
Again, the \(KE\) terms cancel out, leading to:
\[ y = \frac{m_{\alpha} q_p}{m_p q_{\alpha}} \cdot \sqrt{\frac{m_p}{m_{\alpha}}} \]

### Final Ratios
1. The ratio of magnitudes of acceleration \(x\) can be simplified based on the mass and charge ratios.
2. The ratio of time periods \(y\) can also be simplified similarly.