A target element A is bombarded with electrons and the wavelengths of the characterstic spectrum are measured. A second characteristic spectrum is also obtained , because of an impurity in the target. The wavelengths of the `K_(alpha)` lines are 196 pm (element A) and 169 pm (impurity). If the atomic number of impurity is `z=(10lambda-1)` . Find the value of `lambda` . (atomic number of element A is 27).

Answer

To solve the problem, we will use Moseley's law, which relates the wavelength of the characteristic X-rays emitted by an atom to its atomic number. The law states that the wavelength (λ) is inversely proportional to the square of the atomic number (Z) of the element.

### Step-by-Step Solution:

1. **Understanding Moseley's Law**:
According to Moseley's law, the wavelength of the Kα line can be expressed as:
\[ \lambda = \frac{K}{Z - b^2} \]
where \( K \) is a constant and \( b \) is a constant that can be taken as 1 for Kα lines.

2. **Setting Up the Equations**:
For element A with atomic number \( Z_A = 27 \):
\[ \lambda_1 = \frac{K}{27 - 1^2} = \frac{K}{26} \]
For the impurity with atomic number \( Z = 10\lambda - 1 \):
\[ \lambda_2 = \frac{K}{(10\lambda - 1) - 1^2} = \frac{K}{10\lambda - 2} \]

3. **Taking the Ratio of Wavelengths**:
The ratio of the wavelengths for element A and the impurity is given by:
\[ \frac{\lambda_1}{\lambda_2} = \frac{K/26}{K/(10\lambda - 2)} \]
Simplifying this gives:
\[ \frac{\lambda_1}{\lambda_2} = \frac{10\lambda - 2}{26} \]

4. **Substituting the Given Wavelengths**:
We know:
\[ \lambda_1 = 196 \text{ pm} = 196 \times 10^{-12} \text{ m} \]
\[ \lambda_2 = 169 \text{ pm} = 169 \times 10^{-12} \text{ m} \]
Thus, we can write:
\[ \frac{196 \times 10^{-12}}{169 \times 10^{-12}} = \frac{10\lambda - 2}{26} \]
The \( 10^{-12} \) cancels out:
\[ \frac{196}{169} = \frac{10\lambda - 2}{26} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 196 \cdot 26 = 169 \cdot (10\lambda - 2) \]

6. **Calculating the Left Side**:
Calculate \( 196 \cdot 26 \):
\[ 196 \cdot 26 = 5096 \]

7. **Expanding the Right Side**:
Expanding the right side:
\[ 1690\lambda - 338 \]

8. **Setting the Equation**:
Now we have:
\[ 5096 = 1690\lambda - 338 \]

9. **Solving for \( \lambda \)**:
Rearranging gives:
\[ 1690\lambda = 5096 + 338 \]
\[ 1690\lambda = 5434 \]
\[ \lambda = \frac{5434}{1690} \approx 3.22 \]

10. **Final Answer**:
Rounding gives \( \lambda \approx 3 \).

### Conclusion:
The value of \( \lambda \) is approximately **3**.