A proton and an `alpha-`particle enter a uniform magnetic field moving with the same speed. If the proton takes `25 mu s` to make 5 revolutions, then the periodic time for the `alpha-` particle would be

Options

A. `5mu` sec
B. `8mu` sec
C. `10mu` sec
D. `16mu` sec

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the relationship between the time period, mass, and charge of the particles.
The time period \( T \) for a charged particle moving in a magnetic field is given by the formula:
\[ T \propto \frac{m}{qB} \]
where:
- \( m \) is the mass of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength (which is constant for both particles).

### Step 2: Identify the known values for the proton.
From the problem, we know:
- The time taken by the proton for 5 revolutions is \( 25 \, \mu s \).
- Therefore, the time period \( T_p \) for the proton is:
\[ T_p = \frac{25 \, \mu s}{5} = 5 \, \mu s \]

### Step 3: Determine the mass and charge of the proton and alpha particle.
- Mass of proton \( m_p = 1 \, \text{amu} \)
- Charge of proton \( q_p = 1e \)

- Mass of alpha particle \( m_{\alpha} = 4 \, \text{amu} \)
- Charge of alpha particle \( q_{\alpha} = 2e \)

### Step 4: Set up the ratio of the time periods for the proton and alpha particle.
Using the relationship established in Step 1:
\[ \frac{T_{\alpha}}{T_p} = \frac{m_{\alpha}/q_{\alpha}}{m_p/q_p} \]

Substituting the values:
\[ \frac{T_{\alpha}}{T_p} = \frac{4 \, \text{amu} / 2e}{1 \, \text{amu} / 1e} = \frac{4/2}{1/1} = \frac{4}{2} = 2 \]

### Step 5: Calculate the time period for the alpha particle.
Since \( T_{\alpha} = 2 \times T_p \):
\[ T_{\alpha} = 2 \times 5 \, \mu s = 10 \, \mu s \]

### Final Answer:
The periodic time for the alpha particle is \( 10 \, \mu s \).

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