Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?

Options

A. Electron
B. Proton
C. `He^+`
D. `Li^(++)`

Answer

To determine which particle will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:

\[ r = \frac{mv}{qB} \]

Where:
- \( r \) is the radius of the circular path,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step-by-Step Solution:

1. **Understand the Formula**: The radius \( r \) is directly proportional to the mass \( m \) of the particle and the velocity \( v \), and inversely proportional to the charge \( q \) and the magnetic field \( B \). Since \( v \), \( q \), and \( B \) are constant for this problem, we can simplify our analysis to focus on the mass of the particles.

2. **List the Particles and Their Masses**:
- **Electron**: Mass \( m_e \approx 9.11 \times 10^{-31} \) kg
- **Proton**: Mass \( m_p \approx 1.67 \times 10^{-27} \) kg
- **Helium Positive Ion (He\(^+\))**: Mass \( m_{He^+} \approx 4 \times m_p \approx 6.68 \times 10^{-27} \) kg
- **Helium Two Positive Ion (He\(^{++}\))**: Mass \( m_{He^{++}} \approx 4 \times m_p \approx 6.68 \times 10^{-27} \) kg

3. **Compare the Masses**: The electron has the smallest mass compared to the proton and the helium ions. The smaller the mass, the smaller the radius of the circular path, since \( r \) is directly proportional to \( m \).

4. **Conclusion**: Since the electron has the smallest mass, it will describe the smallest circle when projected with the same velocity perpendicular to the magnetic field.

### Final Answer:
The particle that will describe the smallest circle is the **electron**.