Mixed `He^(+)` and `O^(2+)` ions (mass of `He^(+)`=4 amu and that of `O^(2+)=16` amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same then

Options

A. `He^(+)` ions will be deflected more than those of `O^(2+)`
B. `He^(+)` ions will be deflected less than those of `O^(2+)`
C. All the ions will be deflected equally
D. No ions will be deflected

Answer

To solve the problem, we need to analyze the motion of the ions in a magnetic field and how their radii of curvature relate to their kinetic energy and mass.

### Step-by-step Solution:

1. **Understanding the Motion of Charged Particles in a Magnetic Field**:
When a charged particle moves in a magnetic field, it experiences a centripetal force that causes it to move in a circular path. The radius \( R \) of this circular path is given by the formula:
\[ R = \frac{mv}{qB} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength.

2. **Relating Kinetic Energy to Velocity**:
The kinetic energy \( KE \) of a particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
From this, we can express the velocity \( v \) in terms of kinetic energy:
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

3. **Substituting Velocity into the Radius Formula**:
We can substitute the expression for \( v \) into the radius formula:
\[ R = \frac{m \sqrt{\frac{2 \cdot KE}{m}}}{qB} = \frac{\sqrt{2m \cdot KE}}{qB} \]

4. **Calculating the Radius for Each Ion**:
- For \( He^+ \) (mass \( m_{He} = 4 \, \text{amu} \), charge \( q_{He} = e \)):
\[ R_{He} = \frac{\sqrt{2 \cdot 4 \cdot KE}}{eB} = \frac{\sqrt{8 \cdot KE}}{eB} \]
- For \( O^{2+} \) (mass \( m_{O} = 16 \, \text{amu} \), charge \( q_{O} = 2e \)):
\[ R_{O} = \frac{\sqrt{2 \cdot 16 \cdot KE}}{2eB} = \frac{\sqrt{32 \cdot KE}}{2eB} = \frac{\sqrt{32 \cdot KE}}{2eB} \]

5. **Finding the Ratio of the Radii**:
Now, we can find the ratio of the radii of the two ions:
\[ \frac{R_{He}}{R_{O}} = \frac{\sqrt{8 \cdot KE}}{eB} \div \frac{\sqrt{32 \cdot KE}}{2eB} = \frac{\sqrt{8 \cdot KE} \cdot 2eB}{\sqrt{32 \cdot KE} \cdot eB} \]
Simplifying this gives:
\[ \frac{R_{He}}{R_{O}} = \frac{2\sqrt{8}}{\sqrt{32}} = \frac{2\sqrt{8}}{4} = \frac{\sqrt{8}}{2} = 1 \]

6. **Conclusion**:
Since \( R_{He} = R_{O} \), both ions will have the same radius of curvature in the magnetic field, indicating that they will be deflected equally.

### Final Answer:
The radii of curvature for \( He^+ \) and \( O^{2+} \) are equal, meaning they will be deflected equally in the magnetic field.