` H^(+), He^(+) and O^(++)` all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity . The masses of ` H^(+), He^(+) and O^(2+)` are `1 amu, 4 amu and 16 amu` respectively . Then

Options

A. `H^(+)` ions will be deflected most
B. `O^(2+)` ions will be deflected least
C. `He^(+)` and `O^(2+)` ions will suffer same deflection
D. All ions will suffer the same deflection

Answer

To solve the problem, we need to analyze the motion of the ions \( H^+ \), \( He^+ \), and \( O^{2+} \) in a magnetic field. Given that they all have the same kinetic energy and are subjected to a uniform magnetic field that is perpendicular to their velocity, we can derive the radius of their circular paths.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) of a charged particle is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass and \( v \) is the velocity of the particle.

2. **Setting Up the Magnetic Force**:
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F = qvB \]
where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength.

3. **Centripetal Force**:
The magnetic force acts as the centripetal force that keeps the particle moving in a circular path:
\[ F = \frac{mv^2}{R} \]
where \( R \) is the radius of the circular path.

4. **Equating Forces**:
Setting the magnetic force equal to the centripetal force, we have:
\[ qvB = \frac{mv^2}{R} \]
Rearranging gives:
\[ R = \frac{mv}{qB} \]

5. **Expressing Velocity in Terms of Kinetic Energy**:
From the kinetic energy equation, we can express \( v \):
\[ v = \sqrt{\frac{2KE}{m}} \]
Substituting this into the radius formula:
\[ R = \frac{m \sqrt{\frac{2KE}{m}}}{qB} = \frac{\sqrt{2m \cdot KE}}{qB} \]

6. **Calculating the Radius for Each Ion**:
Since all ions have the same kinetic energy, we can analyze the radius for each ion:
- For \( H^+ \) (mass = 1 amu, charge = +1):
\[ R_H = \frac{\sqrt{2 \cdot 1 \cdot KE}}{1B} = \sqrt{\frac{2KE}{B^2}} \]
- For \( He^+ \) (mass = 4 amu, charge = +1):
\[ R_{He} = \frac{\sqrt{2 \cdot 4 \cdot KE}}{1B} = 2\sqrt{\frac{2KE}{B^2}} \]
- For \( O^{2+} \) (mass = 16 amu, charge = +2):
\[ R_O = \frac{\sqrt{2 \cdot 16 \cdot KE}}{2B} = 2\sqrt{\frac{2KE}{B^2}} \]

7. **Comparing the Radii**:
- The radius for \( H^+ \) is \( R_H = \sqrt{\frac{2KE}{B^2}} \)
- The radius for \( He^+ \) is \( R_{He} = 2\sqrt{\frac{2KE}{B^2}} \)
- The radius for \( O^{2+} \) is also \( R_O = 2\sqrt{\frac{2KE}{B^2}} \)

8. **Conclusion**:
The radius of the path for \( H^+ \) is the smallest, while both \( He^+ \) and \( O^{2+} \) have the same larger radius. Thus, \( H^+ \) will experience the maximum deflection, while \( He^+ \) and \( O^{2+} \) will have the same deflection.