Assertion An `alpha-`particle and a deuteron having same kinetic energy enter in a uniform magnetic field perpendicular to the field. Then, radius of circular path of `alpha-`particle will be more.
Reason `(q)/(m)` ratio of an `alpha-`particle is more than the `(q)/(m)` ratio of a deuteron.

Options

A. (a)If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
B. (b)If both Assertion and Reason are true but Reason is not correct explanation of Assertion.
C. (c)If Assertion is false and Reason is false.
D. (d)If Assertion is false but Reason is true.

Answer

To solve the problem, we need to analyze the motion of an alpha particle and a deuteron in a magnetic field, given that they have the same kinetic energy. The radius of the circular path in a magnetic field is determined by the charge-to-mass ratio of the particles.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Both the alpha particle and the deuteron have the same kinetic energy (K.E.).
- They enter a uniform magnetic field (B) perpendicular to the field.

2. **Understand the Motion in a Magnetic Field:**
- 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 the circular path is given by the formula:
\[ r = \frac{mv}{qB} \]
where:
- \( m \) = mass of the particle
- \( v \) = velocity of the particle
- \( q \) = charge of the particle
- \( B \) = magnetic field strength

3. **Relate Kinetic Energy to Velocity:**
- The kinetic energy (K.E.) of a particle is given by:
\[ K.E. = \frac{1}{2}mv^2 \]
- Rearranging this gives:
\[ v = \sqrt{\frac{2 \cdot K.E.}{m}} \]

4. **Substitute Velocity into the Radius Formula:**
- Substitute \( v \) in the radius formula:
\[ r = \frac{m \cdot \sqrt{\frac{2 \cdot K.E.}{m}}}{qB} \]
- This simplifies to:
\[ r = \frac{\sqrt{2 \cdot K.E. \cdot m}}{qB} \]

5. **Compare the Radii of the Alpha Particle and Deuteron:**
- Let \( r_{\alpha} \) be the radius for the alpha particle and \( r_{d} \) for the deuteron.
- The ratio of the radii can be expressed as:
\[ \frac{r_{\alpha}}{r_{d}} = \frac{\sqrt{2 \cdot K.E. \cdot m_{\alpha}} / q_{\alpha}}{\sqrt{2 \cdot K.E. \cdot m_{d}} / q_{d}} = \frac{q_{d} \cdot \sqrt{m_{\alpha}}}{q_{\alpha} \cdot \sqrt{m_{d}}} \]

6. **Evaluate the Charge-to-Mass Ratios:**
- For an alpha particle:
- Charge \( q_{\alpha} = 2e \) (two protons)
- Mass \( m_{\alpha} \approx 4u \) (approximately four atomic mass units)
- For a deuteron:
- Charge \( q_{d} = e \) (one proton)
- Mass \( m_{d} \approx 2u \) (approximately two atomic mass units)
- Thus, the charge-to-mass ratios are:
\[ \frac{q_{\alpha}}{m_{\alpha}} = \frac{2e}{4u} = \frac{e}{2u} \]
\[ \frac{q_{d}}{m_{d}} = \frac{e}{2u} \]

7. **Conclusion:**
- Since the charge-to-mass ratio of the alpha particle is less than that of the deuteron, the radius of the circular path of the alpha particle will be less than that of the deuteron.
- Therefore, the assertion is **false** and the reason is **true**.

### Final Answer:
- Assertion: False
- Reason: True