Deuteron and `alpha`-particle are put `1 A^(@)` apart in Air. Magnitude of intensity of electric field due to deuteron at `alpha`-particle is (N/C).

Options

A. Zero
B. `2.88 xx 10^(11)`
C. `1.44 xx 10^(11)`
D. `5.76 xx 10^(11)`

Answer

To find the magnitude of the intensity of the electric field due to a deuteron at the position of an alpha particle, we can follow these steps:

### Step 1: Identify the charges involved
The charge of a deuteron (which is a hydrogen isotope) is equal to the charge of a proton, which is approximately \( +1e \) or \( +1.6 \times 10^{-19} \) coulombs.

### Step 2: Determine the distance
The distance between the deuteron and the alpha particle is given as \( 1 \, \text{Å} \) (angstrom), which is equal to \( 1 \times 10^{-10} \, \text{m} \).

### Step 3: Use the formula for electric field
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r^2} \]
where:
- \( E \) is the electric field,
- \( Q \) is the charge creating the electric field (in this case, the charge of the deuteron),
- \( r \) is the distance from the charge to the point where the electric field is being calculated,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 4: Substitute values into the formula
Substituting the known values into the formula:
- \( Q = 1.6 \times 10^{-19} \, \text{C} \)
- \( r = 1 \times 10^{-10} \, \text{m} \)
- \( \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

The electric field \( E \) due to the deuteron at the position of the alpha particle is:
\[ E = \left( 9 \times 10^9 \right) \cdot \frac{1.6 \times 10^{-19}}{(1 \times 10^{-10})^2} \]

### Step 5: Calculate the electric field
Calculating the denominator:
\[ (1 \times 10^{-10})^2 = 1 \times 10^{-20} \]
Now substituting back:
\[ E = 9 \times 10^9 \cdot \frac{1.6 \times 10^{-19}}{1 \times 10^{-20}} = 9 \times 10^9 \cdot 1.6 \times 10^{1} = 9 \times 1.6 \times 10^{10} = 14.4 \times 10^{10} \]
Thus, we have:
\[ E = 1.44 \times 10^{11} \, \text{N/C} \]

### Final Answer
The magnitude of the intensity of the electric field due to the deuteron at the position of the alpha particle is:
\[ \boxed{1.44 \times 10^{11} \, \text{N/C}} \]