The energy levels of a hypothetical one - electron atom system are given by `E_n = 16/(n^2) eV`, where `n = 1, 2, 3,…….` . The wavelength of emitted photom corresponding to the transitition from first excited level to ground level is about

Options

A. 1035 Å
B. 1220 Å
C. 3650 Å
D. 690 Å

Answer

To solve the problem, we need to find the wavelength of the emitted photon corresponding to the transition from the first excited level (n=2) to the ground level (n=1) in a hypothetical one-electron atom system with energy levels given by \( E_n = \frac{16}{n^2} \) eV.

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
- For the ground state (n=1):
\[ E_1 = \frac{16}{1^2} = 16 \text{ eV} \]
- For the first excited state (n=2):
\[ E_2 = \frac{16}{2^2} = \frac{16}{4} = 4 \text{ eV} \]

2. **Calculate the Energy Difference**:
- The energy difference \( \Delta E \) when the electron transitions from n=2 to n=1 is given by:
\[ \Delta E = E_1 - E_2 = 16 \text{ eV} - 4 \text{ eV} = 12 \text{ eV} \]

3. **Use the Energy-Wavelength Relation**:
- The energy of a photon is related to its wavelength by the formula:
\[ E = \frac{hc}{\lambda} \]
- Rearranging this gives:
\[ \lambda = \frac{hc}{E} \]
- Alternatively, using the shortcut formula for energy in eV:
\[ \lambda (\text{in angstrom}) = \frac{12420}{E (\text{in eV})} \]

4. **Substituting the Energy**:
- Now substituting \( E = 12 \text{ eV} \) into the shortcut formula:
\[ \lambda = \frac{12420}{12} \]

5. **Calculating the Wavelength**:
- Performing the division:
\[ \lambda = 1035 \text{ angstrom} \]

### Final Answer:
The wavelength of the emitted photon corresponding to the transition from the first excited level to the ground level is approximately **1035 angstrom**.