For a cartain metal, the `K` obsorption edge is at `0.72 Å`. The wavelength of `K_(alpha), K_(beta). And K_(gamma)` lines of of `K` series are `0.210 Å , 0.192 Å, and 0.180 Å, respectively. The eneggies of `K, L and M` orbit are `E_(K), E_(L) and E_(M)`, respectively. Then

Options

A. `E_(K) = - 13.04 keV`
B. `E_(L) = - 7.52 keV`
C. `E_(M) = - 3.21 keV`
D. `E_(K) = - 13.04 keV`

Explanation

To solve the problem, we need to calculate the energies of the K, L, and M orbits based on the given absorption edge and the wavelengths of the K series lines. Here's a step-by-step solution:

### Step 1: Calculate the Energy of the K Absorption Edge
The energy corresponding to the K absorption edge can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \, \text{eV s} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength in meters.

Given that the K absorption edge is at \( 0.72 \, \text{Å} = 0.072 \, \text{nm} \), we convert this to nanometers:
\[ \lambda = 0.072 \, \text{nm} \]
Now, substituting the values into the equation:
\[ E_K = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{0.072 \, \text{nm}} = \frac{1240 \, \text{eV nm}}{0.072 \, \text{nm}} \approx 72.09 \, \text{keV} \]

### Step 2: Calculate the Energy of Kα, Kβ, and Kγ Lines
Using the same formula for the Kα, Kβ, and Kγ lines:

1. **For Kα (0.210 Å = 0.0210 nm)**:
\[ E_{K\alpha} = \frac{1240 \, \text{eV nm}}{0.0210 \, \text{nm}} \approx 59.04 \, \text{keV} \]

2. **For Kβ (0.192 Å = 0.0192 nm)**:
\[ E_{K\beta} = \frac{1240 \, \text{eV nm}}{0.0192 \, \text{nm}} \approx 64.58 \, \text{keV} \]

3. **For Kγ (0.180 Å = 0.0180 nm)**:
\[ E_{K\gamma} = \frac{1240 \, \text{eV nm}}{0.0180 \, \text{nm}} \approx 68.88 \, \text{keV} \]

### Step 3: Calculate the Energies of the K, L, and M Orbits
Now we can find the energies of the K, L, and M orbits using the previously calculated energies:

1. **Energy of K shell (\( E_K \))**:
\[ E_K = E_{K\alpha} - E_K = 59.04 \, \text{keV} - 72.09 \, \text{keV} = -13.05 \, \text{keV} \]

2. **Energy of L shell (\( E_L \))**:
\[ E_L = E_{K\beta} - E_K = 64.58 \, \text{keV} - 72.09 \, \text{keV} = -7.51 \, \text{keV} \]

3. **Energy of M shell (\( E_M \))**:
\[ E_M = E_{K\gamma} - E_K = 68.88 \, \text{keV} - 72.09 \, \text{keV} = -3.21 \, \text{keV} \]

### Summary of Results
- \( E_K \approx -13.05 \, \text{keV} \)
- \( E_L \approx -7.51 \, \text{keV} \)
- \( E_M \approx -3.21 \, \text{keV} \)

### Final Answer
The energies of the K, L, and M orbits are approximately:
- \( E_K = -13.05 \, \text{keV} \)
- \( E_L = -7.51 \, \text{keV} \)
- \( E_M = -3.21 \, \text{keV} \)