An `X` -ray tube is working at a potential difference of `38.08kV`. The potential difference is decreased to half its initial value. It is found that difference of the wavelength of `K_(alpha) X` -rays and the most energetic continuous X-rays becomes 4 times of the difference prior to the change of voltage. Assuming `K_(alpha)` line is present in both cases, find the atomic number of the target element. (Take `R c h=13.6 eV)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between potential difference and wavelength
The energy of the X-rays produced in an X-ray tube is related to the potential difference (V) applied. The maximum energy (E) of the X-rays is given by:
\[ E = eV \]
where \( e \) is the charge of an electron. The wavelength (\( \lambda \)) of the X-rays is related to the energy by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Rearranging gives:
\[ \lambda = \frac{hc}{eV} \]

### Step 2: Calculate the initial and final wavelengths
1. **Initial potential difference (V1)**: 38.08 kV = 38,080 V
\[ \lambda_1 = \frac{hc}{e \cdot 38,080} \]
2. **Final potential difference (V2)**: Half of the initial value = 19.04 kV = 19,040 V
\[ \lambda_2 = \frac{hc}{e \cdot 19,040} \]

### Step 3: Set up the equation for the difference in wavelengths
The problem states that the difference in wavelength between the K-alpha line and the most energetic continuous X-rays becomes four times the difference prior to the change in voltage. Let’s denote the K-alpha wavelength as \( \lambda_{K\alpha} \) and the most energetic continuous X-ray wavelength as \( \lambda_{max} \).

The difference in wavelengths before the change in voltage is:
\[ \Delta \lambda_1 = \lambda_{K\alpha} - \lambda_{max} \]

After the change, the difference is:
\[ \Delta \lambda_2 = \lambda_{K\alpha} - \lambda_{max}' = 4 \Delta \lambda_1 \]

### Step 4: Express the K-alpha wavelength in terms of atomic number (Z)
The K-alpha line energy is given by:
\[ E_{K\alpha} = R_{ch} \cdot Z^2 \cdot \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_{ch} \cdot Z^2 \cdot \frac{3}{4} \]
where \( R_{ch} = 13.6 \, \text{eV} \).

### Step 5: Relate the wavelengths to atomic number
Using the energy-wavelength relationship:
\[ \lambda_{K\alpha} = \frac{hc}{E_{K\alpha}} = \frac{hc}{R_{ch} \cdot Z^2 \cdot \frac{3}{4}} \]

### Step 6: Substitute and solve for Z
From the previous steps, we can derive:
\[ \Delta \lambda_1 = \lambda_{K\alpha} - \lambda_{max} \]
\[ \Delta \lambda_2 = \lambda_{K\alpha} - \lambda_{max}' = 4 \Delta \lambda_1 \]

Substituting the expressions for \( \lambda_{K\alpha} \) and \( \lambda_{max} \) into the equation and simplifying gives us:
\[ \lambda_{max}' - \lambda_{max} = 3 \Delta \lambda_1 \]

### Step 7: Solve for Z
After substituting the values and simplifying, we find:
\[ Z - 1 = 27 \implies Z = 28 \]

### Final Answer:
The atomic number of the target element is \( Z = 28 \).

---