If an electron undergoes transition from n = 2 to n =1 in `Li^(2+)` ion , the energy of photon radiated will be best given by

Options

A. hv
B. `h v_(1) + h v_(2)`
C. `hv_(1) + hv_(2) + hv_(3)`
D. All of these

Explanation

To find the energy of the photon emitted when an electron transitions from n = 2 to n = 1 in the lithium ion \( Li^{2+} \), we can use the formula for the energy levels of hydrogen-like atoms. The energy levels are given by the formula:

\[ E_n = -\frac{Z^2 \cdot R_H}{n^2} \]

where:
- \( E_n \) is the energy of the level,
- \( Z \) is the atomic number (for \( Li^{2+} \), \( Z = 3 \)),
- \( R_H \) is the Rydberg constant (approximately \( 13.6 \, eV \)),
- \( n \) is the principal quantum number.

### Step 1: Calculate the energy of the levels

1. **Calculate \( E_2 \)** (for n = 2):
\[ E_2 = -\frac{3^2 \cdot 13.6}{2^2} = -\frac{9 \cdot 13.6}{4} = -30.6 \, eV \]

2. **Calculate \( E_1 \)** (for n = 1):
\[ E_1 = -\frac{3^2 \cdot 13.6}{1^2} = -9 \cdot 13.6 = -122.4 \, eV \]

### Step 2: Find the energy difference

3. **Calculate the energy difference \( \Delta E \)** between the two levels:
\[ \Delta E = E_1 - E_2 = (-122.4) - (-30.6) = -122.4 + 30.6 = -91.8 \, eV \]

### Step 3: Photon energy

4. **The energy of the photon emitted** when the electron transitions from n = 2 to n = 1 is the absolute value of the energy difference:
\[ E_{photon} = |\Delta E| = 91.8 \, eV \]

### Final Answer

The energy of the photon radiated when an electron transitions from n = 2 to n = 1 in \( Li^{2+} \) is approximately **91.8 eV**.

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