Determine the energy of the characteristic X-ray `(K_beta)` emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1)

Explanation

To determine the energy of the characteristic X-ray (K_beta) emitted from a tungsten (Z = 74) target when an electron drops from the M-shell (n=3) to a vacancy in the K-shell (n=1), we can follow these steps:

### Step 1: Calculate the energy of the K-shell (E_K)
The energy of an electron in the K-shell (n=1) can be calculated using the formula:

\[ E_K = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]

For tungsten (Z = 74):

\[ E_K = -\frac{74^2 \cdot 13.6}{1^2} = -\frac{5476 \cdot 13.6}{1} = -74 \cdot 13.6 = -1013.6 \, \text{eV} \]

### Step 2: Calculate the effective nuclear charge (Z_eff) for the M-shell
The effective nuclear charge experienced by the electron in the M-shell can be calculated by considering the shielding effect. The number of electrons in the inner shells (K and L shells) is 9 (2 in K and 8 in L). Thus, we have:

\[ Z_{\text{eff}} = Z - \text{shielding electrons} = 74 - 9 = 65 \]

### Step 3: Calculate the energy of the M-shell (E_M)
Using the effective nuclear charge, we calculate the energy of the electron in the M-shell (n=3):

\[ E_M = -\frac{Z_{\text{eff}}^2 \cdot 13.6 \, \text{eV}}{n^2} \]

Substituting the values:

\[ E_M = -\frac{65^2 \cdot 13.6}{3^2} = -\frac{4225 \cdot 13.6}{9} = -\frac{57400}{9} \approx -6388.89 \, \text{eV} \]

### Step 4: Calculate the energy of the emitted X-ray (E_X-ray)
The energy of the emitted X-ray when an electron transitions from the M-shell to the K-shell is given by the difference in energy between these two states:

\[ E_X-ray = E_M - E_K \]

Substituting the values we calculated:

\[ E_X-ray = (-6388.89) - (-1013.6) = -6388.89 + 1013.6 = -5375.29 \, \text{eV} \]

### Step 5: Final energy of the K_beta X-ray
The energy of the K_beta X-ray emitted is approximately:

\[ E_X-ray \approx 5375.29 \, \text{eV} \]

### Final Answer:
The energy of the characteristic X-ray (K_beta) emitted from a tungsten target is approximately **5375.29 eV**.

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