In an X-ray experiment, target is made up of copper `(Z=29)` having some impurity. The `K_(alpha)` line of copper have wavelength `lambda_(0)`. It was observed that another `K_(alpha)` line due to impurity have wavelength `(784)/(625) lambda_(0)`. The atomic number of the impurity element is

Answer

To find the atomic number of the impurity element in the X-ray experiment, we can follow these steps:

### Step 1: Write the formula for the wavelength of the K-alpha line
The wavelength of the K-alpha line for an element with atomic number \( Z \) is given by the formula:
\[ \lambda_{K_{\alpha}} = \frac{1216}{(Z - 1)^2} \text{ angstroms} \]

### Step 2: Set up the relationship between the wavelengths
We know that the wavelength of the K-alpha line for copper is \( \lambda_0 \) and for the impurity is given as:
\[ \lambda_{impurity} = \frac{784}{625} \lambda_0 \]

### Step 3: Write the equations for both wavelengths
For copper (with \( Z = 29 \)):
\[ \lambda_0 = \frac{1216}{(29 - 1)^2} = \frac{1216}{28^2} \]

For the impurity (let's denote its atomic number as \( Z \)):
\[ \lambda_{impurity} = \frac{1216}{(Z - 1)^2} \]

### Step 4: Substitute the expression for \( \lambda_{impurity} \)
Substituting the expression for \( \lambda_{impurity} \) into the equation:
\[ \frac{784}{625} \lambda_0 = \frac{1216}{(Z - 1)^2} \]

### Step 5: Substitute \( \lambda_0 \) into the equation
Substituting \( \lambda_0 = \frac{1216}{28^2} \):
\[ \frac{784}{625} \cdot \frac{1216}{28^2} = \frac{1216}{(Z - 1)^2} \]

### Step 6: Simplify the equation
Cancelling \( 1216 \) from both sides:
\[ \frac{784}{625 \cdot 28^2} = \frac{1}{(Z - 1)^2} \]

### Step 7: Cross-multiply to solve for \( Z - 1 \)
Cross-multiplying gives:
\[ (Z - 1)^2 = \frac{625 \cdot 28^2}{784} \]

### Step 8: Calculate \( 28^2 \) and simplify
Calculating \( 28^2 = 784 \):
\[ (Z - 1)^2 = \frac{625 \cdot 784}{784} = 625 \]

### Step 9: Take the square root
Taking the square root of both sides:
\[ Z - 1 = 25 \]

### Step 10: Solve for \( Z \)
Adding 1 to both sides gives:
\[ Z = 25 + 1 = 26 \]

Thus, the atomic number of the impurity element is \( \boxed{26} \).

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