A deuteron and an `alpha`-particle are placed in an electric field.The forces acting on them are `F_(1)` and `F_(2)` and their accelerations are `a_(1)` and `a_(2)` respectively.
(i) `F_(1) = F_(2)`
(ii) `F_(1) != F_(2)`
(iii) `a_(1) = a_(2)`
(iv) `a_(1) != a_(2)`

Options

A. `F_(1) = F_(2)`
B. `F_(1) gt F_(2)`
C. `a_(1) = a_(2)`
D. `a_(1) cancel(=) a_(2)`

Answer

To solve the problem, we need to analyze the forces and accelerations acting on a deuteron and an alpha particle placed in an electric field.

### Step-by-Step Solution:

1. **Identify the Charges:**
- A deuteron consists of one proton and one neutron, giving it a charge of \( +e \) (where \( e \) is the elementary charge).
- An alpha particle consists of two protons and two neutrons, giving it a charge of \( +2e \).

2. **Determine the Forces:**
- The force \( F \) acting on a charged particle in an electric field \( E \) is given by the equation:
\[ F = qE \]
- For the deuteron:
\[ F_1 = q_1 E = eE \]
- For the alpha particle:
\[ F_2 = q_2 E = 2eE \]

3. **Compare the Forces:**
- Since \( F_1 = eE \) and \( F_2 = 2eE \), we can see that:
\[ F_1 \neq F_2 \]
- Therefore, the forces acting on the deuteron and the alpha particle are not equal.

4. **Determine the Masses:**
- The mass of a deuteron is approximately \( 2m \) (where \( m \) is the mass of a proton).
- The mass of an alpha particle is approximately \( 4m \).

5. **Calculate the Accelerations:**
- Using Newton's second law \( F = ma \), we can express the accelerations:
\[ a_1 = \frac{F_1}{m_1} = \frac{eE}{2m} \]
\[ a_2 = \frac{F_2}{m_2} = \frac{2eE}{4m} = \frac{eE}{2m} \]
- From the above equations, we see that:
\[ a_1 = a_2 \]
- Therefore, the accelerations of the deuteron and the alpha particle are equal.

### Conclusion:
- From the analysis, we conclude:
- (ii) \( F_1 \neq F_2 \) (the forces are not equal)
- (iii) \( a_1 = a_2 \) (the accelerations are equal)

### Final Answer:
The correct options are (ii) and (iii).