An alpha particle and a deuteron are accelerated in a cyclotron. If `K_(1)" and " K_(2)` be the maximum kinetic energy acquired by the alpha particle and the deuteron, respectively, then `K_(1)//K_(2)` would be:

Options

A. 2
B. `sqrt2`
C. 1
D. 4

Explanation

To solve the problem, we need to determine the ratio of the maximum kinetic energy acquired by an alpha particle (K1) and a deuteron (K2) when they are accelerated in a cyclotron.

### Step-by-Step Solution:

1. **Understanding the Particles**:
- An alpha particle (α) is a helium nucleus, which consists of 2 protons and 2 neutrons. Thus, its charge (q1) is \(2e\) and its mass (m1) is approximately \(4m\) (where \(m\) is the mass of a nucleon).
- A deuteron (D) is an isotope of hydrogen, consisting of 1 proton and 1 neutron. Its charge (q2) is \(e\) and its mass (m2) is approximately \(2m\).

2. **Kinetic Energy in a Cyclotron**:
- The kinetic energy (K) of a charged particle in a cyclotron can be expressed as:
\[ K = \frac{1}{2} mv^2 \]
- The velocity \(v\) of a particle in a cyclotron is given by:
\[ v = \frac{qBr}{m} \]
- Substituting this expression for \(v\) into the kinetic energy formula gives:
\[ K = \frac{1}{2} m \left(\frac{qBr}{m}\right)^2 = \frac{q^2 B^2 r^2}{2m} \]

3. **Calculating Kinetic Energies**:
- For the alpha particle:
\[ K_1 = \frac{(2e)^2 B^2 r^2}{2 \cdot 4m} = \frac{4e^2 B^2 r^2}{8m} = \frac{e^2 B^2 r^2}{2m} \]
- For the deuteron:
\[ K_2 = \frac{(e)^2 B^2 r^2}{2 \cdot 2m} = \frac{e^2 B^2 r^2}{4m} \]

4. **Finding the Ratio \(K_1/K_2\)**:
- Now, we can find the ratio of the kinetic energies:
\[ \frac{K_1}{K_2} = \frac{\frac{e^2 B^2 r^2}{2m}}{\frac{e^2 B^2 r^2}{4m}} = \frac{4}{2} = 2 \]

5. **Conclusion**:
- Therefore, the ratio of the maximum kinetic energies acquired by the alpha particle and the deuteron is:
\[ \frac{K_1}{K_2} = 2 \]

### Final Answer:
\[ \frac{K_1}{K_2} = 2 \]