The energy of the n=2 state in a given element is `E_2 =-2870 eV`. Given that the wavelengths of lthe `K_alpha` and `K_beta` lines are 0.71 nm and 0.63 nm respectively, determine the energies `E_1` and `E_3`.

Explanation

To solve the problem, we need to determine the energies \( E_1 \) and \( E_3 \) based on the given information about the energy of the \( n=2 \) state and the wavelengths of the \( K_\alpha \) and \( K_\beta \) lines.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Energy of the \( n=2 \) state: \( E_2 = -2870 \, \text{eV} \)
- Wavelength of \( K_\alpha \) line: \( \lambda_{K_\alpha} = 0.71 \, \text{nm} = 0.71 \times 10^{-9} \, \text{m} \)
- Wavelength of \( K_\beta \) line: \( \lambda_{K_\beta} = 0.63 \, \text{nm} = 0.63 \times 10^{-9} \, \text{m} \)

2. **Use the Energy-Wavelength Relation:**
The energy difference corresponding to a transition can be calculated using the formula:
\[ \Delta E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant \( (6.63 \times 10^{-34} \, \text{Js}) \) and \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \).

3. **Calculate \( E_1 \) using \( K_\alpha \):**
The transition for \( K_\alpha \) occurs from \( E_2 \) to \( E_1 \):
\[ E_2 - E_1 = \Delta E_{K_\alpha} \]
Rearranging gives:
\[ E_1 = E_2 - \Delta E_{K_\alpha} \]
Now calculate \( \Delta E_{K_\alpha} \):
\[ \Delta E_{K_\alpha} = \frac{hc}{\lambda_{K_\alpha}} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{0.71 \times 10^{-9}} \]
Calculating this gives:
\[ \Delta E_{K_\alpha} \approx 2.79 \, \text{keV} = 2790 \, \text{eV} \]
Now substitute back to find \( E_1 \):
\[ E_1 = -2870 \, \text{eV} - 2790 \, \text{eV} = -5660 \, \text{eV} \]

4. **Calculate \( E_3 \) using \( K_\beta \):**
The transition for \( K_\beta \) occurs from \( E_3 \) to \( E_1 \):
\[ E_3 - E_1 = \Delta E_{K_\beta} \]
Rearranging gives:
\[ E_3 = E_1 + \Delta E_{K_\beta} \]
Now calculate \( \Delta E_{K_\beta} \):
\[ \Delta E_{K_\beta} = \frac{hc}{\lambda_{K_\beta}} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{0.63 \times 10^{-9}} \]
Calculating this gives:
\[ \Delta E_{K_\beta} \approx 3.14 \, \text{keV} = 3140 \, \text{eV} \]
Now substitute back to find \( E_3 \):
\[ E_3 = -5660 \, \text{eV} + 3140 \, \text{eV} = -2520 \, \text{eV} \]

### Final Results:
- \( E_1 \approx -5660 \, \text{eV} \)
- \( E_3 \approx -2520 \, \text{eV} \)