`A` and `B` are two hydrogen like atoms such that `m_(B)=2m_(A)`. Also, the number of protons and neutrons in the two nuclei are equal. Given that difference of photon energy corresponding to the first Balmer lines emitted by `A` and `B` is `2.667eV`. Let `Z_(A)` and `Z_(B)` be the atomic numbers of `A` and `B` respectively.

Options

A. `Z_(A)=4`
B. `Z_(A)=1`
C. `Z_(B)=2`
D. `Z_(B)=8`

Explanation

To solve the problem step by step, we will analyze the given information and use the relevant formulas for hydrogen-like atoms.

### Step 1: Understand the relationship between the atomic numbers and masses.
Given that \( m_B = 2m_A \) and the number of protons and neutrons in the two nuclei are equal, we can infer that the atomic number \( Z_B \) of atom B is related to the atomic number \( Z_A \) of atom A. Since the mass of atom B is twice that of atom A, we can assume that the atomic number of B is also twice that of A.

**Hint:** Remember that for hydrogen-like atoms, the atomic number is directly related to the mass of the nucleus.

### Step 2: Write the energy difference formula for the first Balmer line.
The energy of the photon emitted during the transition to the second energy level (first Balmer line) for hydrogen-like atoms is given by:
\[ E_n = \frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
For the first Balmer line (transition from n=3 to n=2), the energy difference for atoms A and B can be expressed as:
\[ E_B - E_A = \left( \frac{13.6 \cdot Z_B^2}{4} \right) - \left( \frac{13.6 \cdot Z_A^2}{4} \right) \]

**Hint:** Use the formula for energy levels of hydrogen-like atoms and remember that the energy difference is proportional to the square of the atomic number.

### Step 3: Set up the equation for the energy difference.
The difference in energy is given as \( 2.667 \, \text{eV} \):
\[ E_B - E_A = \frac{13.6}{4} (Z_B^2 - Z_A^2) = 2.667 \]

**Hint:** Factor out the common terms to simplify the equation.

### Step 4: Substitute \( Z_B = 2Z_A \) into the equation.
Substituting \( Z_B = 2Z_A \) into the energy difference equation:
\[ \frac{13.6}{4} \left( (2Z_A)^2 - Z_A^2 \right) = 2.667 \]
This simplifies to:
\[ \frac{13.6}{4} \left( 4Z_A^2 - Z_A^2 \right) = 2.667 \]
\[ \frac{13.6}{4} \cdot 3Z_A^2 = 2.667 \]

**Hint:** Simplifying the equation will help you isolate \( Z_A^2 \).

### Step 5: Solve for \( Z_A^2 \).
Now, we can solve for \( Z_A^2 \):
\[ 3Z_A^2 = \frac{2.667 \cdot 4}{13.6} \]
Calculating the right side:
\[ 3Z_A^2 = \frac{10.668}{13.6} \approx 0.785 \]
\[ Z_A^2 = \frac{0.785}{3} \approx 0.2617 \]

**Hint:** Take the square root to find \( Z_A \).

### Step 6: Calculate \( Z_A \) and \( Z_B \).
Taking the square root:
\[ Z_A \approx 0.511 \approx 1 \quad (\text{since atomic numbers are whole numbers}) \]
Now, substituting back to find \( Z_B \):
\[ Z_B = 2Z_A = 2 \cdot 1 = 2 \]

**Hint:** Verify the values against the conditions given in the problem.

### Final Result:
Thus, the atomic numbers are:
\[ Z_A = 1, \quad Z_B = 2 \]