Given that in H-atom the transitioin energy for `n=1` to n=2 is 10.2eV, the energy for the same transition in `Be^(3+)` is

Options

A. 20.4eV
B. 30.6eV
C. 40.8eV
D. None of these

Answer

To find the energy for the transition from \( n=1 \) to \( n=2 \) in the \( \text{Be}^{3+} \) ion, we can use the formula for the energy levels of hydrogen-like atoms:

\[ \Delta E = 13.6 \, \text{eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step-by-Step Solution:

**Step 1: Identify the values for \( Z \), \( n_1 \), and \( n_2 \)**
For \( \text{Be}^{3+} \):
- The atomic number \( Z \) of beryllium (Be) is 4.
- The principal quantum numbers are \( n_1 = 1 \) and \( n_2 = 2 \).

**Step 2: Substitute the values into the formula**
Now, we substitute these values into the energy transition formula:

\[ \Delta E = 13.6 \, \text{eV} \times 4^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

**Step 3: Calculate \( Z^2 \)**
Calculating \( Z^2 \):

\[ Z^2 = 4^2 = 16 \]

**Step 4: Calculate the fraction**
Now calculate the fraction:

\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

**Step 5: Substitute back into the equation**
Now substitute \( Z^2 \) and the fraction back into the equation:

\[ \Delta E = 13.6 \, \text{eV} \times 16 \times \frac{3}{4} \]

**Step 6: Simplify the expression**
Calculating the expression:

\[ \Delta E = 13.6 \, \text{eV} \times 16 \times 0.75 \]

**Step 7: Calculate the final value**
Now calculate:

\[ \Delta E = 13.6 \times 12 = 163.2 \, \text{eV} \]

### Final Answer:
The energy for the transition from \( n=1 \) to \( n=2 \) in \( \text{Be}^{3+} \) is \( 163.2 \, \text{eV} \).

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